{"id":1149,"date":"2021-11-11T17:37:24","date_gmt":"2021-11-11T17:37:24","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/skills-review-for-conic-sections\/"},"modified":"2022-11-09T16:27:11","modified_gmt":"2022-11-09T16:27:11","slug":"skills-review-for-conic-sections","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/skills-review-for-conic-sections\/","title":{"raw":"Skills Review for Conic Sections","rendered":"Skills Review for Conic Sections"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Find the distance between two points<\/li>\r\n \t<li>Find the midpoint of a line segment<\/li>\r\n \t<li>Complete the square<\/li>\r\n<\/ul>\r\n<\/div>\r\nIn the Conic Section section, we will learn about various conic sections including parabolas, ellipses, and hyperbolas. Here we will review how to use both the distance and midpoint formulas along with how to complete the square.\r\n<h2>Find the Distance Between Two Points<\/h2>\r\nDerived from the <strong>Pythagorean Theorem<\/strong>, the <strong>distance formula<\/strong> is used to find the distance between two points in the plane. The Pythagorean Theorem, [latex]{a}^{2}+{b}^{2}={c}^{2}[\/latex], is based on a right triangle where <em>a <\/em>and <em>b<\/em> are the lengths of the legs adjacent to the right angle, and <em>c<\/em> is the length of the hypotenuse.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/12042428\/CNX_CAT_Figure_02_01_015.jpg\" alt=\"This is an image of a triangle on an x, y coordinate plane. The x and y axes range from 0 to 7. The points (x sub 1, y sub 1); (x sub 2, y sub 1); and (x sub 2, y sub 2) are labeled and connected to form a triangle. Along the base of the triangle, the following equation is displayed: the absolute value of x sub 2 minus x sub 1 equals a. The hypotenuse of the triangle is labeled: d = c. The remaining side is labeled: the absolute value of y sub 2 minus y sub 1 equals b.\" width=\"487\" height=\"331\" \/>\r\n\r\nThe relationship of sides [latex]|{x}_{2}-{x}_{1}|[\/latex] and [latex]|{y}_{2}-{y}_{1}|[\/latex] to side <em>d<\/em> is the same as that of sides <em>a <\/em>and <em>b <\/em>to side <em>c.<\/em> We use the absolute value symbol to indicate that the length is a positive number because the absolute value of any number is positive. (For example, [latex]|-3|=3[\/latex]. ) The symbols [latex]|{x}_{2}-{x}_{1}|[\/latex] and [latex]|{y}_{2}-{y}_{1}|[\/latex] indicate that the lengths of the sides of the triangle are positive. To find the length <em>c<\/em>, take the square root of both sides of the Pythagorean Theorem.\r\n<div style=\"text-align: center;\">[latex]{c}^{2}={a}^{2}+{b}^{2}\\rightarrow c=\\sqrt{{a}^{2}+{b}^{2}}[\/latex]<\/div>\r\nIt follows that the distance formula is given as\r\n<div style=\"text-align: center;\">[latex]{d}^{2}={\\left({x}_{2}-{x}_{1}\\right)}^{2}+{\\left({y}_{2}-{y}_{1}\\right)}^{2}\\to d=\\sqrt{{\\left({x}_{2}-{x}_{1}\\right)}^{2}+{\\left({y}_{2}-{y}_{1}\\right)}^{2}}[\/latex]<\/div>\r\nWe do not have to use the absolute value symbols in this definition because any number squared is positive.\r\n<div class=\"textbox\">\r\n<h3>A General Note: The Distance Formula<\/h3>\r\nGiven endpoints [latex]\\left({x}_{1},{y}_{1}\\right)[\/latex] and [latex]\\left({x}_{2},{y}_{2}\\right)[\/latex], the distance between two points is given by\r\n<div style=\"text-align: center;\">[latex]d=\\sqrt{{\\left({x}_{2}-{x}_{1}\\right)}^{2}+{\\left({y}_{2}-{y}_{1}\\right)}^{2}}[\/latex]<\/div>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Finding the Distance between Two Points<\/h3>\r\nFind the distance between the points [latex]\\left(-3,-1\\right)[\/latex] and [latex]\\left(2,3\\right)[\/latex].\r\n\r\n[reveal-answer q=\"737169\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"737169\"]\r\n\r\nLet us first look at the graph of the two points. Connect the points to form a right triangle.\r\n\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/12042430\/CNX_CAT_Figure_02_01_016.jpg\" alt=\"This is an image of a triangle on an x, y coordinate plane. The x-axis ranges from negative 4 to 4. The y-axis ranges from negative 2 to 4. The points (-3, -1); (2, -1); and (2, 3) are plotted and labeled on the graph. The points are connected to form a triangle\" width=\"487\" height=\"289\" \/>\r\n\r\nThen, calculate the length of <em>d <\/em>using the distance formula.\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}d=\\sqrt{{\\left({x}_{2}-{x}_{1}\\right)}^{2}+{\\left({y}_{2}-{y}_{1}\\right)}^{2}}\\hfill \\\\ d=\\sqrt{{\\left(2-\\left(-3\\right)\\right)}^{2}+{\\left(3-\\left(-1\\right)\\right)}^{2}}\\hfill \\\\ =\\sqrt{{\\left(5\\right)}^{2}+{\\left(4\\right)}^{2}}\\hfill \\\\ =\\sqrt{25+16}\\hfill \\\\ =\\sqrt{41}\\hfill \\end{array}[\/latex]<\/div>\r\n<div>[\/hidden-answer]<\/div>\r\n<\/div>\r\n<div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<div>\r\n\r\nFind the distance between two points: [latex]\\left(1,4\\right)[\/latex] and [latex]\\left(11,9\\right)[\/latex].\r\n[reveal-answer q=\"934526\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"934526\"]\r\n\r\n[latex]\\sqrt{125}=5\\sqrt{5}[\/latex][\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<h2>Find the Midpoint of a Line Segment<\/h2>\r\nWhen the endpoints of a line segment are known, we can find the point midway between them. This point is known as the midpoint and the formula is known as the <strong>midpoint formula<\/strong>. Given the endpoints of a line segment, [latex]\\left({x}_{1},{y}_{1}\\right)[\/latex] and [latex]\\left({x}_{2},{y}_{2}\\right)[\/latex], the midpoint formula states how to find the coordinates of the midpoint [latex]M[\/latex].\r\n<div style=\"text-align: center;\">[latex]M=\\left(\\frac{{x}_{1}+{x}_{2}}{2},\\frac{{y}_{1}+{y}_{2}}{2}\\right)[\/latex]<\/div>\r\nA graphical view of a midpoint is shown below. Notice that the line segments on either side of the midpoint are congruent.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/12042436\/CNX_CAT_Figure_02_01_018.jpg\" alt=\"This is a line graph on an x, y coordinate plane with the x and y axes ranging from 0 to 6. The points (x sub 1, y sub 1), (x sub 2, y sub 2), and (x sub 1 plus x sub 2 all over 2, y sub 1 plus y sub 2 all over 2) are plotted. A straight line runs through these three points. Pairs of short parallel lines bisect the two sections of the line to note that they are equivalent.\" width=\"487\" height=\"290\" \/>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Finding the Midpoint of A Line Segment<\/h3>\r\nFind the midpoint of the line segment with the endpoints [latex]\\left(7,-2\\right)[\/latex] and [latex]\\left(9,5\\right)[\/latex].\r\n\r\n[reveal-answer q=\"788934\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"788934\"]\r\n\r\nUse the formula to find the midpoint of the line segment.\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}\\left(\\frac{{x}_{1}+{x}_{2}}{2},\\frac{{y}_{1}+{y}_{2}}{2}\\right)\\hfill&amp;=\\left(\\frac{7+9}{2},\\frac{-2+5}{2}\\right)\\hfill \\\\ \\hfill&amp;=\\left(8,\\frac{3}{2}\\right)\\hfill \\end{array}[\/latex]<\/div>\r\n<div>[\/hidden-answer]<\/div>\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nFind the midpoint of the line segment with endpoints [latex]\\left(-2,-1\\right)[\/latex] and [latex]\\left(-8,6\\right)[\/latex].\r\n\r\n[reveal-answer q=\"964077\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"964077\"]\r\n\r\n[latex]\\left(-5,\\frac{5}{2}\\right)[\/latex][\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<iframe id=\"mom2\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=2308&amp;theme=oea&amp;iframe_resize_id=mom2\" width=\"100%\" height=\"250\" data-mce-fragment=\"1\"><\/iframe>\r\n\r\n<\/div>\r\n<h2>Complete the Square<\/h2>\r\nWe will use the example [latex]{x}^{2}+4x+1=0[\/latex] to illustrate each step of completing the sqare. To complete the square, it is always easier when the leading coefficient, <em>a<\/em>, equals 1.\r\n<ol>\r\n \t<li>Given a quadratic equation that cannot be factored and with [latex]a=1[\/latex], first add or subtract the constant term to the right sign of the equal sign.\r\n<div style=\"text-align: center;\">[latex]{x}^{2}+4x=-1[\/latex]<\/div><\/li>\r\n \t<li>Multiply the <em>b <\/em>term by [latex]\\frac{1}{2}[\/latex] and square it.\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}\\frac{1}{2}\\left(4\\right)=2\\hfill \\\\ {2}^{2}=4\\hfill \\end{array}[\/latex]<\/div><\/li>\r\n \t<li>Add [latex]{\\left(\\frac{1}{2}b\\right)}^{2}[\/latex] to both sides of the equal sign and simplify the right side. We have\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}{x}^{2}+4x+4=-1+4\\hfill \\\\ {x}^{2}+4x+4=3\\hfill \\end{array}[\/latex]<\/div><\/li>\r\n \t<li>The left side of the equation can now be factored as a perfect square.\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}{x}^{2}+4x+4=3\\hfill \\\\ {\\left(x+2\\right)}^{2}=3\\hfill \\end{array}[\/latex]<\/div><\/li>\r\n \t<li>Use the square root property and solve.\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}\\sqrt{{\\left(x+2\\right)}^{2}}=\\pm \\sqrt{3}\\hfill \\\\ x+2=\\pm \\sqrt{3}\\hfill \\\\ x=-2\\pm \\sqrt{3}\\hfill \\end{array}[\/latex]<\/div><\/li>\r\n \t<li>The solutions are [latex]x=-2+\\sqrt{3}[\/latex], [latex]x=-2-\\sqrt{3}[\/latex].<\/li>\r\n<\/ol>\r\nWe will see when studying conic sections that the method of completing the square comes in handy when rewriting the equation of a conic section given in general form. In preparation to use the method in this manner, it will be good to practice it first.\r\n<div class=\"textbox\">\r\n<h3>How To: use the method of complete the square to write a perfect square trinomial from an expression<\/h3>\r\n<ol>\r\n \t<li>Given an expression of the form [latex]a\\left(x^2+bx\\right)[\/latex], add [latex]\\left(\\dfrac{b}{2}\\right)^2[\/latex] inside the parentheses.<\/li>\r\n \t<li>Then subtract [latex]a\\left(\\dfrac{b}{2}\\right)^2[\/latex] to counteract the change you made to the expression.<\/li>\r\n \t<li>If completing the square on one side of an equation, you may either subtract the value of\u00a0[latex]a\\left(\\dfrac{b}{2}\\right)^2[\/latex] from that side, or add it to the other to maintain equality.<\/li>\r\n \t<li>\u00a0Then factor the perfect square trinomial you created inside the original parentheses.<\/li>\r\n<\/ol>\r\n<h3>The resulting form will look like this:<\/h3>\r\nGiven\r\n<p style=\"text-align: center;\">[latex]\\qquad a\\left(x^2+bx\\right)[\/latex]<\/p>\r\nadd [latex]\\left(b\/2\\right)^2[\/latex] inside the parentheses and subtract\u00a0[latex]a\\left(b\/2\\right)^2[\/latex] to counteract the change you made to the expression\r\n<p style=\"text-align: center;\">[latex]=a\\left(x^2+bx+ \\left(\\dfrac{b}{2}\\right)^2\\right)-a\\left(\\dfrac{b}{2}\\right)^2[\/latex]<\/p>\r\nthen factor the resulting perfect square trinomial\r\n<p style=\"text-align: center;\">[latex]=a\\left(x+ \\dfrac{b}{2}\\right)^2-a\\left(\\dfrac{b}{2}\\right)^2[\/latex].<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example : CreatING a perfect square trinomial using the method of complete the square<\/h3>\r\nComplete the square on: [latex]3\\left(x^2 - 10x\\right)[\/latex].\r\n\r\n[reveal-answer q=\"107002\"]Show Solution[\/reveal-answer]\r\n\r\n[hidden-answer a=\"107002\"]\r\n\r\n&nbsp;\r\n\r\nAdd [latex]\\left(\\dfrac{b}{2}\\right)^2[\/latex] inside the parentheses and subtract [latex]a\\left(\\dfrac{b}{2}\\right)^2[\/latex]\r\n<p style=\"text-align: center;\">[latex]3\\left(x^2 - 10x+25\\right)-3\\cdot25[\/latex]<\/p>\r\n&nbsp;\r\n\r\nFactor the perfect square trinomial and simplify\r\n<p style=\"text-align: center;\">[latex]3\\left(x -5\\right)^2-75[\/latex].<\/p>\r\n.\r\n<h4><span style=\"text-decoration: underline;\">Analysis<\/span><\/h4>\r\nThe resulting expression is equivalent to the original expression. To test this, substitute a small value for [latex]x[\/latex], say [latex]x=3[\/latex].\r\n\r\n[latex]3\\left(3^2-10\\cdot3\\right) = \\quad -63 \\quad = 3(3-5)^2-75[\/latex]. True.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\n[ohm_question]15002[\/ohm_question]\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Find the distance between two points<\/li>\n<li>Find the midpoint of a line segment<\/li>\n<li>Complete the square<\/li>\n<\/ul>\n<\/div>\n<p>In the Conic Section section, we will learn about various conic sections including parabolas, ellipses, and hyperbolas. Here we will review how to use both the distance and midpoint formulas along with how to complete the square.<\/p>\n<h2>Find the Distance Between Two Points<\/h2>\n<p>Derived from the <strong>Pythagorean Theorem<\/strong>, the <strong>distance formula<\/strong> is used to find the distance between two points in the plane. The Pythagorean Theorem, [latex]{a}^{2}+{b}^{2}={c}^{2}[\/latex], is based on a right triangle where <em>a <\/em>and <em>b<\/em> are the lengths of the legs adjacent to the right angle, and <em>c<\/em> is the length of the hypotenuse.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/12042428\/CNX_CAT_Figure_02_01_015.jpg\" alt=\"This is an image of a triangle on an x, y coordinate plane. The x and y axes range from 0 to 7. The points (x sub 1, y sub 1); (x sub 2, y sub 1); and (x sub 2, y sub 2) are labeled and connected to form a triangle. Along the base of the triangle, the following equation is displayed: the absolute value of x sub 2 minus x sub 1 equals a. The hypotenuse of the triangle is labeled: d = c. The remaining side is labeled: the absolute value of y sub 2 minus y sub 1 equals b.\" width=\"487\" height=\"331\" \/><\/p>\n<p>The relationship of sides [latex]|{x}_{2}-{x}_{1}|[\/latex] and [latex]|{y}_{2}-{y}_{1}|[\/latex] to side <em>d<\/em> is the same as that of sides <em>a <\/em>and <em>b <\/em>to side <em>c.<\/em> We use the absolute value symbol to indicate that the length is a positive number because the absolute value of any number is positive. (For example, [latex]|-3|=3[\/latex]. ) The symbols [latex]|{x}_{2}-{x}_{1}|[\/latex] and [latex]|{y}_{2}-{y}_{1}|[\/latex] indicate that the lengths of the sides of the triangle are positive. To find the length <em>c<\/em>, take the square root of both sides of the Pythagorean Theorem.<\/p>\n<div style=\"text-align: center;\">[latex]{c}^{2}={a}^{2}+{b}^{2}\\rightarrow c=\\sqrt{{a}^{2}+{b}^{2}}[\/latex]<\/div>\n<p>It follows that the distance formula is given as<\/p>\n<div style=\"text-align: center;\">[latex]{d}^{2}={\\left({x}_{2}-{x}_{1}\\right)}^{2}+{\\left({y}_{2}-{y}_{1}\\right)}^{2}\\to d=\\sqrt{{\\left({x}_{2}-{x}_{1}\\right)}^{2}+{\\left({y}_{2}-{y}_{1}\\right)}^{2}}[\/latex]<\/div>\n<p>We do not have to use the absolute value symbols in this definition because any number squared is positive.<\/p>\n<div class=\"textbox\">\n<h3>A General Note: The Distance Formula<\/h3>\n<p>Given endpoints [latex]\\left({x}_{1},{y}_{1}\\right)[\/latex] and [latex]\\left({x}_{2},{y}_{2}\\right)[\/latex], the distance between two points is given by<\/p>\n<div style=\"text-align: center;\">[latex]d=\\sqrt{{\\left({x}_{2}-{x}_{1}\\right)}^{2}+{\\left({y}_{2}-{y}_{1}\\right)}^{2}}[\/latex]<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Finding the Distance between Two Points<\/h3>\n<p>Find the distance between the points [latex]\\left(-3,-1\\right)[\/latex] and [latex]\\left(2,3\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q737169\">Show Solution<\/span><\/p>\n<div id=\"q737169\" class=\"hidden-answer\" style=\"display: none\">\n<p>Let us first look at the graph of the two points. Connect the points to form a right triangle.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/12042430\/CNX_CAT_Figure_02_01_016.jpg\" alt=\"This is an image of a triangle on an x, y coordinate plane. The x-axis ranges from negative 4 to 4. The y-axis ranges from negative 2 to 4. The points (-3, -1); (2, -1); and (2, 3) are plotted and labeled on the graph. The points are connected to form a triangle\" width=\"487\" height=\"289\" \/><\/p>\n<p>Then, calculate the length of <em>d <\/em>using the distance formula.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}d=\\sqrt{{\\left({x}_{2}-{x}_{1}\\right)}^{2}+{\\left({y}_{2}-{y}_{1}\\right)}^{2}}\\hfill \\\\ d=\\sqrt{{\\left(2-\\left(-3\\right)\\right)}^{2}+{\\left(3-\\left(-1\\right)\\right)}^{2}}\\hfill \\\\ =\\sqrt{{\\left(5\\right)}^{2}+{\\left(4\\right)}^{2}}\\hfill \\\\ =\\sqrt{25+16}\\hfill \\\\ =\\sqrt{41}\\hfill \\end{array}[\/latex]<\/div>\n<div><\/div>\n<\/div>\n<\/div>\n<\/div>\n<div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<div>\n<p>Find the distance between two points: [latex]\\left(1,4\\right)[\/latex] and [latex]\\left(11,9\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q934526\">Show Solution<\/span><\/p>\n<div id=\"q934526\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\sqrt{125}=5\\sqrt{5}[\/latex]<\/p><\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<h2>Find the Midpoint of a Line Segment<\/h2>\n<p>When the endpoints of a line segment are known, we can find the point midway between them. This point is known as the midpoint and the formula is known as the <strong>midpoint formula<\/strong>. Given the endpoints of a line segment, [latex]\\left({x}_{1},{y}_{1}\\right)[\/latex] and [latex]\\left({x}_{2},{y}_{2}\\right)[\/latex], the midpoint formula states how to find the coordinates of the midpoint [latex]M[\/latex].<\/p>\n<div style=\"text-align: center;\">[latex]M=\\left(\\frac{{x}_{1}+{x}_{2}}{2},\\frac{{y}_{1}+{y}_{2}}{2}\\right)[\/latex]<\/div>\n<p>A graphical view of a midpoint is shown below. Notice that the line segments on either side of the midpoint are congruent.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/12042436\/CNX_CAT_Figure_02_01_018.jpg\" alt=\"This is a line graph on an x, y coordinate plane with the x and y axes ranging from 0 to 6. The points (x sub 1, y sub 1), (x sub 2, y sub 2), and (x sub 1 plus x sub 2 all over 2, y sub 1 plus y sub 2 all over 2) are plotted. A straight line runs through these three points. Pairs of short parallel lines bisect the two sections of the line to note that they are equivalent.\" width=\"487\" height=\"290\" \/><\/p>\n<div class=\"textbox exercises\">\n<h3>Example: Finding the Midpoint of A Line Segment<\/h3>\n<p>Find the midpoint of the line segment with the endpoints [latex]\\left(7,-2\\right)[\/latex] and [latex]\\left(9,5\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q788934\">Show Solution<\/span><\/p>\n<div id=\"q788934\" class=\"hidden-answer\" style=\"display: none\">\n<p>Use the formula to find the midpoint of the line segment.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}\\left(\\frac{{x}_{1}+{x}_{2}}{2},\\frac{{y}_{1}+{y}_{2}}{2}\\right)\\hfill&=\\left(\\frac{7+9}{2},\\frac{-2+5}{2}\\right)\\hfill \\\\ \\hfill&=\\left(8,\\frac{3}{2}\\right)\\hfill \\end{array}[\/latex]<\/div>\n<div><\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Find the midpoint of the line segment with endpoints [latex]\\left(-2,-1\\right)[\/latex] and [latex]\\left(-8,6\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q964077\">Show Solution<\/span><\/p>\n<div id=\"q964077\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\left(-5,\\frac{5}{2}\\right)[\/latex]<\/p><\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"mom2\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=2308&amp;theme=oea&amp;iframe_resize_id=mom2\" width=\"100%\" height=\"250\" data-mce-fragment=\"1\"><\/iframe><\/p>\n<\/div>\n<h2>Complete the Square<\/h2>\n<p>We will use the example [latex]{x}^{2}+4x+1=0[\/latex] to illustrate each step of completing the sqare. To complete the square, it is always easier when the leading coefficient, <em>a<\/em>, equals 1.<\/p>\n<ol>\n<li>Given a quadratic equation that cannot be factored and with [latex]a=1[\/latex], first add or subtract the constant term to the right sign of the equal sign.\n<div style=\"text-align: center;\">[latex]{x}^{2}+4x=-1[\/latex]<\/div>\n<\/li>\n<li>Multiply the <em>b <\/em>term by [latex]\\frac{1}{2}[\/latex] and square it.\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}\\frac{1}{2}\\left(4\\right)=2\\hfill \\\\ {2}^{2}=4\\hfill \\end{array}[\/latex]<\/div>\n<\/li>\n<li>Add [latex]{\\left(\\frac{1}{2}b\\right)}^{2}[\/latex] to both sides of the equal sign and simplify the right side. We have\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}{x}^{2}+4x+4=-1+4\\hfill \\\\ {x}^{2}+4x+4=3\\hfill \\end{array}[\/latex]<\/div>\n<\/li>\n<li>The left side of the equation can now be factored as a perfect square.\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}{x}^{2}+4x+4=3\\hfill \\\\ {\\left(x+2\\right)}^{2}=3\\hfill \\end{array}[\/latex]<\/div>\n<\/li>\n<li>Use the square root property and solve.\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}\\sqrt{{\\left(x+2\\right)}^{2}}=\\pm \\sqrt{3}\\hfill \\\\ x+2=\\pm \\sqrt{3}\\hfill \\\\ x=-2\\pm \\sqrt{3}\\hfill \\end{array}[\/latex]<\/div>\n<\/li>\n<li>The solutions are [latex]x=-2+\\sqrt{3}[\/latex], [latex]x=-2-\\sqrt{3}[\/latex].<\/li>\n<\/ol>\n<p>We will see when studying conic sections that the method of completing the square comes in handy when rewriting the equation of a conic section given in general form. In preparation to use the method in this manner, it will be good to practice it first.<\/p>\n<div class=\"textbox\">\n<h3>How To: use the method of complete the square to write a perfect square trinomial from an expression<\/h3>\n<ol>\n<li>Given an expression of the form [latex]a\\left(x^2+bx\\right)[\/latex], add [latex]\\left(\\dfrac{b}{2}\\right)^2[\/latex] inside the parentheses.<\/li>\n<li>Then subtract [latex]a\\left(\\dfrac{b}{2}\\right)^2[\/latex] to counteract the change you made to the expression.<\/li>\n<li>If completing the square on one side of an equation, you may either subtract the value of\u00a0[latex]a\\left(\\dfrac{b}{2}\\right)^2[\/latex] from that side, or add it to the other to maintain equality.<\/li>\n<li>\u00a0Then factor the perfect square trinomial you created inside the original parentheses.<\/li>\n<\/ol>\n<h3>The resulting form will look like this:<\/h3>\n<p>Given<\/p>\n<p style=\"text-align: center;\">[latex]\\qquad a\\left(x^2+bx\\right)[\/latex]<\/p>\n<p>add [latex]\\left(b\/2\\right)^2[\/latex] inside the parentheses and subtract\u00a0[latex]a\\left(b\/2\\right)^2[\/latex] to counteract the change you made to the expression<\/p>\n<p style=\"text-align: center;\">[latex]=a\\left(x^2+bx+ \\left(\\dfrac{b}{2}\\right)^2\\right)-a\\left(\\dfrac{b}{2}\\right)^2[\/latex]<\/p>\n<p>then factor the resulting perfect square trinomial<\/p>\n<p style=\"text-align: center;\">[latex]=a\\left(x+ \\dfrac{b}{2}\\right)^2-a\\left(\\dfrac{b}{2}\\right)^2[\/latex].<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example : CreatING a perfect square trinomial using the method of complete the square<\/h3>\n<p>Complete the square on: [latex]3\\left(x^2 - 10x\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q107002\">Show Solution<\/span><\/p>\n<div id=\"q107002\" class=\"hidden-answer\" style=\"display: none\">\n<p>&nbsp;<\/p>\n<p>Add [latex]\\left(\\dfrac{b}{2}\\right)^2[\/latex] inside the parentheses and subtract [latex]a\\left(\\dfrac{b}{2}\\right)^2[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]3\\left(x^2 - 10x+25\\right)-3\\cdot25[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p>Factor the perfect square trinomial and simplify<\/p>\n<p style=\"text-align: center;\">[latex]3\\left(x -5\\right)^2-75[\/latex].<\/p>\n<p>.<\/p>\n<h4><span style=\"text-decoration: underline;\">Analysis<\/span><\/h4>\n<p>The resulting expression is equivalent to the original expression. To test this, substitute a small value for [latex]x[\/latex], say [latex]x=3[\/latex].<\/p>\n<p>[latex]3\\left(3^2-10\\cdot3\\right) = \\quad -63 \\quad = 3(3-5)^2-75[\/latex]. True.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm15002\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=15002&theme=oea&iframe_resize_id=ohm15002&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-1149\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Calculus Volume 1. <strong>Provided by<\/strong>: Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/courses.lumenlearning.com\/calculus1\/\">https:\/\/courses.lumenlearning.com\/calculus1\/<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Calculus Volume 2. <strong>Provided by<\/strong>: Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/courses.lumenlearning.com\/calculus2\/\">https:\/\/courses.lumenlearning.com\/calculus2\/<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":349141,"menu_order":4,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Calculus Volume 1\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"https:\/\/courses.lumenlearning.com\/calculus1\/\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Calculus Volume 2\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"https:\/\/courses.lumenlearning.com\/calculus2\/\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-1149","chapter","type-chapter","status-publish","hentry"],"part":1145,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/1149","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/users\/349141"}],"version-history":[{"count":3,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/1149\/revisions"}],"predecessor-version":[{"id":6481,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/1149\/revisions\/6481"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/parts\/1145"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/1149\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/media?parent=1149"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapter-type?post=1149"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/contributor?post=1149"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/license?post=1149"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}