Add 3 to both sides to isolate the variable term on the left side of the equation.

[latex]\begin{array}{r}\sqrt{x}-3\,\,\,=\,\,\,5\\\underline{+3\,\,\,\,\,\,\,+3}\end{array}[/latex]

Collect like terms.

[latex]\sqrt{x}=8[/latex]

Square both sides to remove the radical, since [latex]{{(\sqrt{x})}^{2}}=x[/latex]. Make sure to square the 8 also! Then simplify.

[latex]\begin{array}{r}{{(\sqrt{x})}^{2}}={{8}^{2}}\\x=64\end{array}[/latex]

Answer

[latex]x=64[/latex] is the solution to [latex]\sqrt{x}-3=5[/latex]

Notice how the radical contains a binomial: [latex]x+8[/latex]. Square both sides to remove the radical.

[latex]{{\left( \sqrt{x+8} \right)}^{2}}={{\left( 3 \right)}^{2}}[/latex]

[latex]{{\left( \sqrt{x+8} \right)}^{2}}=x+8[/latex]. Now simplify the equation and solve for x.

[latex]\begin{array}{r}x+8=9\\x=1\end{array}[/latex]

Check your answer. Substituting 1 for x in the original equation yields a true statement, so the solution is correct.

[latex]\begin{array}{r}\sqrt{1+8}=3\\\sqrt{9}=3\\3=3\end{array}[/latex]

Answer

[latex]x=1[/latex] is the solution to [latex]\sqrt{x+8}=3[/latex].

Begin by subtracting 1 from both sides in order to isolate the radical term. Then square both sides to remove the binomial from the radical.

[latex]\begin{array}{r}1+\sqrt{2x+3}-1=6-1\\\sqrt{2x+3}=5\,\,\,\,\,\,\,\,\,\,\\{{\left( \sqrt{2x+3} \right)}^{2}}={{\left( 5 \right)}^{2}}\,\,\,\end{array}[/latex]

Simplify the equation and solve for x.

[latex]\begin{array}{r}2x+3=25\\2x=22\\x=11\end{array}[/latex]

Check your answer. Substituting 11 for x in the original equation yields a true statement, so the solution is correct.

[latex]\begin{array}{r}1+\sqrt{2(11)+3}=6\\1+\sqrt{22+3}=6\\1+\sqrt{25}=6\\1+5=6\\6=6\end{array}[/latex]

Answer

[latex]x=11[/latex] is the solution for [latex]1+\sqrt{2x+3}=6[/latex].

Square both sides to remove the term [latex]a–5[/latex] from the radical.

[latex]{{\left( \sqrt{a-5} \right)}^{2}}={{(-2)}^{2}}[/latex]

Write the simplified equation, and solve for a.

[latex]\begin{array}{r}a-5=4\\a=9\end{array}[/latex]

 

Now check the solution by substituting [latex]a=9[/latex] into the original equation.

It does not check!

[latex]\begin{array}{r}\sqrt{9-5}=-2\\\sqrt{4}=-2\\2\ne -2\end{array}[/latex]

Answer

No solution.

Square both sides to remove the term [latex]x+10[/latex] from the radical.

[latex]{{\left( x+4 \right)}^{2}}={{\left( \sqrt{x+10} \right)}^{2}}[/latex]

Now simplify and solve the equation. Combine like terms, and then factor.

[latex]\begin{array}{r}\left( x+4 \right)\left( x+4 \right)=x+10\\{{x}^{2}}+8x+16=x+10\\{{x}^{2}}+8x-x+16-10=0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\\{{x}^{2}}+7x+6=0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\\\left( x+6 \right)\left( x+1 \right)=0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\end{array}[/latex]

Set each factor equal to zero and solve for x.

[latex]\begin{array}{c}\left( x+6 \right)=0\,\,\text{or}\,\,\left( x+1 \right)=0\\x=-6\text{ or }x=-1\end{array}[/latex]

Now check both solutions by substituting them into the original equation.

Since [latex]x=−6[/latex] produces a false statement, it is an extraneous solution.

[latex]\begin{array}{l}-6+4=\sqrt{-6+10}\\\,\,\,\,\,\,\,\,-2=\sqrt{4}\\\,\,\,\,\,\,\,\,-2=2\\\text{FALSE!}\\\\\\-1+4=\sqrt{-1+10}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3=\sqrt{9}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3=3\\\text{TRUE!}\end{array}[/latex]

Answer

[latex]x=−1[/latex] is the only solution

Isolate the radical term.

[latex]\sqrt{x+2}=x-4[/latex]

Square both sides to remove the term [latex]x+2[/latex] from the radical.

[latex]{{\left( \sqrt{x+2} \right)}^{2}}={{\left( x-4 \right)}^{2}}[/latex]

Now simplify and solve the equation. Combine like terms, and then factor.

[latex]\begin{array}{l}x+2={{x}^{2}}-8x+16\\\,\,\,\,\,\,\,\,\,\,\,0={{x}^{2}}-8x-x+16-2\\\,\,\,\,\,\,\,\,\,\,\,0={{x}^{2}}-9x+14\\\,\,\,\,\,\,\,\,\,\,\,0=\left( x-7 \right)\left( x-2 \right)\end{array}[/latex]

Set each factor equal to zero and solve for x.

[latex]\begin{array}{c}\left( x-7 \right)=0\text{ or }\left( x-2 \right)=0\\x=7\text{ or }x=2\end{array}[/latex][latex][/latex]

Now check both solutions by substituting them into the original equation.

Since [latex]x=2[/latex] produces a false statement, it is an extraneous solution.

[latex]\begin{array}{r}4+\sqrt{7+2}=7\\4+\sqrt{9}=7\\4+3=7\\7=7\\\text{TRUE!}\\\\4+\sqrt{2+2}=2\\4+\sqrt{4}=2\\4+2=2\\6=2\\\text{FALSE!}\end{array}[/latex]

Answer

[latex]x=7[/latex] is the only solution.

Since one side is simply [latex]x^{2}[/latex], you can take the square root of both sides to get x on one side. Don’t forget to use both positive and negative square roots!

[latex]\begin{array}{l}x^{2}=9\\\,\,\,x=\pm\sqrt{9}\\\,\,\,x=\pm3\end{array}[/latex]

Answer

[latex]x=\pm3[/latex] (that is, [latex]x=3[/latex] or [latex]-3[/latex])

If you try taking the square root of both sides of the original equation, you will have [latex]\sqrt{10{{x}^{2}}+5}[/latex] on the left, and you can’t simplify that. Subtract 5 from both sides to get the [latex]x^{2}[/latex] term by itself.

[latex]10x^{2}+5=85[/latex]

You could now take the square root of both sides, but you would have [latex]\sqrt{10}[/latex] as a coefficient, and you would need to divide by that coefficient. Dividing by 10 before you take the square root will be a little easier.

[latex]10x^{2}=80[/latex]

Now you have only [latex]x^{2}[/latex] on the left, so you can use the Square Root Property easily.

Be sure to simplify the radical if possible.

[latex]\begin{array}{l}{{x}^{2}}=8\\\,\,\,x=\pm \sqrt{8}\\\,\,\,\,\,\,=\pm \sqrt{(4)(2)}\\\,\,\,\,\,\,=\pm \sqrt{4}\sqrt{2}\\\,\,\,\,\,\,=\pm 2\sqrt{2}\end{array}[/latex]

Answer

[latex]x=\pm 2\sqrt{2}[/latex]

Again, taking the square root of both sides at this stage will leave something you can’t work with on the left. Start by adding 50 to both sides.

[latex]\left(x-2\right)^{2}-50=0[/latex]

Because [latex]\left(x–2\right)^{2}[/latex] is a squared quantity, you can take the square root of both sides.

[latex]\begin{array}{r}\left(x-2\right)^{2}=50\,\,\,\,\,\,\,\,\,\,\\x-2=\pm\sqrt{50}\end{array}[/latex]

To isolate x on the left, you need to add 2 to both sides.

Be sure to simplify the radical if possible.

[latex]\begin{array}{l}x=2\pm \sqrt{50}\\\,\,\,\,=2\pm \sqrt{(25)(2)}\\\,\,\,\,=2\pm \sqrt{25}\sqrt{2}\\\,\,\,\,=2\pm 5\sqrt{2}\end{array}[/latex]

Answer

[latex]x=2\pm 5\sqrt{2}[/latex]

First notice that the [latex]x^{2}[/latex] term and the constant term are both perfect squares.

[latex]\begin{array}{l}9x^{2}=\left(3x\right)^{2}\\\,\,\,16=4^{2}\end{array}[/latex]

Then notice that the middle term (ignoring the sign) is twice the product of the square roots of the other terms.

[latex]24x=2\left(3x\right)\left(4\right)[/latex]

A trinomial in the form [latex]r^{2}-2rs+s^{2}[/latex] can be factored as [latex](r–s)^{2}[/latex].

In this case, the middle term is subtracted, so subtract r and s and square it to get [latex](r–s)^{2}[/latex].

[latex]\begin{array}{c}\,\,\,r=3x\\s=4\\9x^{2}-24x+16=\left(3x-4\right)^{2}\end{array}[/latex]

Answer

[latex]\left(3x–4\right)^{2}[/latex]

Since there’s an x term, you can’t use the Square Root Property immediately (or even after adding or dividing by a constant).

Notice, however, that the [latex]x^{2}[/latex] and constant terms on the left are both perfect squares: [latex]\left(2x\right)^{2}[/latex] and [latex]5^{2}[/latex]. Check the middle term: is it [latex]2\left(2x\right)\left(5\right)[/latex]? Yes!

[latex]4x^{2}+20x+25=8[/latex]

A trinomial in the form [latex]r^{2}+2rs+s^{2}[/latex] can be factored as [latex]\left(r+s\right)^{2}[/latex], so rewrite the left side as a squared binomial.

[latex](2x+5)^{2}=8[/latex]

Now you can use the Square Root Property. Some additional steps are needed to isolate x.

[latex]\begin{array}{r}2x+5=\pm \sqrt{8}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\\\\2x=-5\pm \sqrt{8}\,\,\,\,\,\\\\x=-\frac{5}{2}\pm \frac{1}{2}\sqrt{8}\end{array}[/latex]

Simplify the radical when possible.

[latex]\begin{array}{l}x=-\frac{5}{2}\pm \frac{1}{2}\sqrt{4}\sqrt{2}\\\\x=-\frac{5}{2}\pm \frac{1}{2}(2)\sqrt{2}\\\\x=-\frac{5}{2}\pm \sqrt{2}\end{array}[/latex]

Answer

[latex]x=-\frac{5}{2}\pm \sqrt{2}[/latex]

First identify b if this has the form [latex]x^{2}+bx[/latex].

[latex]\begin{array}{c}x^{2}+8x\\b=8\end{array}[/latex]

To complete the square, add [latex]\left(\frac{b}{2}\right)^{2}[/latex].

[latex]b=8[/latex], so [latex]\left(\frac{b}{2}\right)^{2}=\left(\frac{8}{2}\right)^{2}[/latex]

Simplify.

[latex]\begin{array}{c}x^{2}+8x+\left(4\right)^{2}\\x^{2}+8x+16\end{array}[/latex]

Check that the result is a perfect square trinomial. [latex]\left(x+4\right)^{2}=x^{2}+4x+4x+16=x^{2}+8x+16[/latex], so it is.

Answer

Adding [latex]+16[/latex] will make [latex]x^{2}+8x[/latex] a perfect square trinomial.

This equation has a constant of 8. Ignore it for now and focus on the [latex]x^{2}[/latex] and x terms on the left side of the equation. The left side has the form [latex]x^{2}+bx[/latex], so you can identify b.

[latex]\begin{array}{r}x^{2}+6x=8\\b=6\end{array}[/latex]

To complete the square, add [latex]{{\left( \frac{b}{2} \right)}^{2}}[/latex] to the left side.

[latex]b=6[/latex], so [latex]{{\left( \frac{b}{2} \right)}^{2}}={{\left( \frac{6}{2} \right)}^{2}}={{3}^{2}}=9.[/latex]

This is an equation, though, so you must add the same number to the right side as well.

[latex]x^{2}+6x+9=8+9[/latex]

Simplify. Check that the left side is a perfect square trinomial. [latex]\begin{array}{r}\left(x+3\right)^{2}=x^{2}+3x+3x+9=x^{2}+6x+9\end{array}{r}[/latex], so it is.

[latex]\begin{array}{r}x^{2}+6x+9=17\\x^{2}+6x+9=17\\(x+3)^{2}=17\end{array}[/latex]

Answer

[latex]x^{2}+6x+9=17[/latex]

Since you cannot factor the trinomial on the left side, you will use completing the square to solve the equation.

Rewrite the equation with the left side in the form [latex]x^{2}+bx[/latex], to prepare to complete the square. Identify b.

[latex]\begin{array}{r}x^{2}-12x=4\,\,\,\,\,\,\,\,\\b=-12\end{array}[/latex]

Figure out what value to add to complete the square. Add [latex]{{\left( \frac{b}{2}\right)}^{2}}[/latex] to complete the square, so [latex]{{\left( \frac{b}{2} \right)}^{2}}={{\left( \frac{-12}{2} \right)}^{2}}={{\left( -6 \right)}^{2}}=36[/latex].

Add the value to both sides of the equation and simplify.

[latex]\begin{array}{l}x^{2}-12x+36=4+36\\x^{2}-12x+36=40\end{array}[/latex]

Rewrite the left side as a squared binomial.

[latex]\left(x-6\right)^{2}=40[/latex]

Use the Square Root Property. Remember to include both the positive and negative square root, or you’ll miss one of the solutions.

[latex]x-6=\pm\sqrt{40}[/latex]

Solve for x by adding 6 to both sides. Simplify as needed.

[latex]\begin{array}{l}x=6\pm \sqrt{40}\\\,\,\,\,=6\pm \sqrt{4}\sqrt{10}\\\,\,\,\,=6\pm 2\sqrt{10}\end{array}[/latex]

Answer

[latex]x=6\pm 2\sqrt{10}[/latex]

Rewrite the equation so the left side has the form [latex]x^{2}+bx[/latex]. Identify b.

[latex]\begin{array}{c}x^{2}+16x=-64\\b=16\end{array}[/latex]

Add [latex]{{\left( \frac{b}{2} \right)}^{2}}[/latex], which is [latex]{{\left( \frac{16}{2} \right)}^{2}}={{8}^{2}}=64[/latex], to both sides.

[latex]\begin{array}{l}x^{2}+16x+64=-64+64\\x^{2}+16x+64=0\end{array}[/latex]

Write the left side as a squared binomial.

[latex]\left(x+8\right)^{2}=0[/latex]

Take the square roots of both sides. Normally both positive and negative square roots are needed, but 0 is neither positive nor negative. 0 has only one root.

[latex]x+8=0[/latex]

Solve for x.

[latex]x=-8[/latex]

Answer

[latex]x=-8[/latex]

First be sure that the right side of the equation is 0. In this case, all you need to do is subtract 5 from both sides.

[latex]\begin{array}{c}3x+2x^{2}+4=5\\3x+2x^{2}+4–5=5–5\end{array}[/latex]

Simplify, and write the terms with the exponent on the variable in descending order.

[latex]\begin{array}{r}3x+2x^{2}-1=0\\2x^{2}+3x-1=0\end{array}[/latex]

Now that the equation is in standard form, you can read the values of a, b, and c from the coefficients and constant. Note that since the constant 1 is subtracted, c must be negative.

[latex]\begin{array}{l}2x^{2}\,\,\,+\,\,\,3x\,\,\,-\,\,\,1\,\,\,=\,\,\,0\\\,\,\downarrow\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\downarrow\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\downarrow\\\,ax^{2}\,\,\,\,\,\,\,\,\,\,bx\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,c\\\\\,\,a=2,\,\,b=3,\,\,c=-1\end{array}[/latex]

Answer

[latex]2x^{2}+3x–1=0;a=2,b=3,c=−1[/latex]

First be sure that the right side of the equation is 0.

[latex]\begin{array}{c}2\left(x+3\right)^{2}–5x=6\\2(x+3)^{2}–5x–6=6–6\end{array}[/latex]

Expand the squared binomial, then simplify by combining like terms.

Be sure to write the terms with the exponent on the variable in descending order.

[latex]\begin{array}{r}2\left(x^{2}+6x+9\right)-5x-6=0\\2x^{2}+12x+18–5x–6=0\\2x^{2}+12x–5x+18–6=0\\2x^{2}+7x+12=0\end{array}[/latex]

Now that the equation is in standard form, you can read the values of a, b, and c from the coefficients and constant.

[latex]\begin{array}{l}2x^{2}\,\,\,+\,\,\,7x\,\,\,+\,\,\,12\,\,\,=\,\,\,0\\\,\downarrow\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\downarrow\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\downarrow\\\,\,a\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,b\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,c\\\\\,\,\,\,\,\,a=2,\,\,b=7,\,\,c=7\end{array}[/latex]

Answer

[latex]2x^{2}+7x+12=0;\,\,a=2,b=7,c=12[/latex]

First write the equation in standard form.

[latex]\begin{array}{r}x^{2}+4x=5\,\,\,\\x^{2}+4x-5=0\,\,\,\\\\a=1,b=4,c=-5\end{array}[/latex]

Note that the subtraction sign means the constant c is negative.

[latex]\begin{array}{r}{{x}^{2}}\,\,\,+\,\,\,4x\,\,\,-\,\,\,5\,\,\,=\,\,\,0\\\downarrow\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\downarrow\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\downarrow\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\\a{{x}^{2}}\,\,\,+\,\,\,bx\,\,\,+\,\,\,c\,\,\,=\,\,\,0\end{array}[/latex]

Substitute the values into the Quadratic Formula. [latex]x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}[/latex]

[latex]\begin{array}{l}\\x=\frac{-4\pm \sqrt{{{(4)}^{2}}-4(1)(-5)}}{2(1)}\end{array}[/latex]

Simplify, being careful to get the signs correct.

[latex]x=\frac{-4\pm\sqrt{16+20}}{2}[/latex]

Simplify some more.

[latex]x=\frac{-4\pm \sqrt{36}}{2}[/latex]

Simplify the radical: [latex]\sqrt{36}=6[/latex].

[latex]x=\frac{-4\pm 6}{2}[/latex]

Separate and simplify to find the solutions to the quadratic equation. Note that in one, 6 is added and in the other, 6 is subtracted.

[latex]\begin{array}{c}x=\frac{-4+6}{2}=\frac{2}{2}=1\\\\\text{or}\\\\x=\frac{-4-6}{2}=\frac{-10}{2}=-5\end{array}[/latex]

Answer

[latex]x=1\,\,\,\text{or}\,\,\,-5[/latex]

Subtract 6x from each side and add 16 to both sides to put the equation in standard form.

[latex]\begin{array}{l}x^{2}-2x=6x-16\\x^{2}-2x-6x+16=0\\x^{2}-8x+16=0\end{array}[/latex]

Identify the coefficients a, b, and c. [latex]x^{2}=1x^{2}[/latex], so [latex]a=1[/latex]. Since [latex]8x[/latex] is subtracted, b is negative. [latex]a=1,b=-8,c=16[/latex]

[latex]\begin{array}{r}{{x}^{2}}\,\,\,-\,\,\,8x\,\,\,+\,\,\,16\,\,\,=\,\,\,0\\\downarrow\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\downarrow\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\downarrow\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\\a{{x}^{2}}\,\,\,+\,\,\,bx\,\,\,+\,\,\,\,c\,\,\,\,=\,\,\,0\end{array}[/latex]

Substitute the values into the Quadratic Formula.

[latex]\begin{array}{l}x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\\\\x=\frac{-(-8)\pm \sqrt{{{(-8)}^{2}}-4(1)(16)}}{2(1)}\end{array}[/latex]

Simplify.

[latex]x=\frac{8\pm \sqrt{64-64}}{2}[/latex]

Since the square root of 0 is 0, and both adding and subtracting 0 give the same result, there is only one possible value.

[latex]x=\frac{8\pm \sqrt{0}}{2}=\frac{8}{2}=4[/latex]

Answer

[latex]x=4[/latex]

When the ball hits the ground, the height is 0. Substitute 0 for h.

[latex]\begin{array}{c}h=-16t^{2}-10t+200\\0=-16t^{2}-10t+200\\-16t^{2}-10t+200=0\end{array}[/latex]

This equation is difficult to solve by factoring or by completing the square, so solve it by applying the Quadratic Formula, [latex]x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}[/latex]. In this case, the variable is t rather than x. [latex]a=−16,b=−10[/latex], and [latex]c=200[/latex].

[latex]t=\frac{-(-10)\pm \sqrt{{{(-10)}^{2}}-4(-16)(200)}}{2(-16)}[/latex]

Simplify. Be very careful with the signs.

[latex]\begin{array}{l}t=\frac{10\pm \sqrt{100+12800}}{-32}\\\,\,=\frac{10\pm \sqrt{12900}}{-32}\end{array}[/latex]

Use a calculator to find both roots.

t is approximately [latex]−3.86[/latex] or [latex]3.24[/latex].

Consider the roots logically. One solution, [latex]−3.86[/latex], cannot be the time because it is a negative number. The other solution, [latex]3.24[/latex] seconds, must be when the ball hits the ground.

Answer

The ball hits the ground approximately [latex]3.24[/latex] seconds after being thrown.

Sketch the problem. Since you don’t know the width of the border, you will let the variable x represent the width.

In the diagram, the original quilt is indicated by the red rectangle. The border is the area between the red and blue lines.

Since each side of the original 4 by 5 quilt has the border of width x added, the length of the quilt with the border will be [latex]5+2x[/latex], and the width will be [latex]4+2x[/latex].

(Both dimensions are written in terms of the same variable, and you will multiply them to get an area! This is where you might start to think that a quadratic equation might be used to solve this problem.)

You are only interested in the area of the border strips. Write an expression for the area of the border.

Area of border = Area of the blue rectangle minus the area of the red rectangle

Area of border[latex]=\left(4+2x\right)\left(5+2x\right)–\left(4\right)\left(5\right)[/latex]

There are 10 sq ft of fabric for the border, so set the area of border to be 10.

[latex]10=\left(4+2x\right)\left(5+2x\right)–20[/latex]

Multiply [latex]\left(4+2x\right)\left(5+2x\right)[/latex].

[latex]10=20+8x+10x+4x^{2}–20[/latex]

Simplify.

[latex]10=18x+4x^{2}[/latex]

Subtract 10 from both sides so that you have a quadratic equation in standard form and can apply the Quadratic Formula to find the roots of the equation.

[latex]\begin{array}{c}0=18x+4x^{2}-10\\\\\text{or}\\\\4x^{2}-10\\\\2\left(2x^{2}+9x-5\right)=0\end{array}[/latex]

Factor out the greatest common factor, 2, so that you can work with the simpler equivalent equation, [latex]2x^{2}+9x–5=0[/latex].

[latex]\begin{array}{r}2\left(2x^{2}+9x-5\right)=0\\\\\frac{2\left(2x^{2}+9x-5\right)}{2}=\frac{0}{2}\\\\2x^{2}+9x-5=0\end{array}[/latex]

Use the Quadratic Formula. In this case, [latex]a=2,b=9[/latex], and [latex]c=−5[/latex].

[latex]\begin{array}{l}x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\\\\x=\frac{-9\pm \sqrt{{{9}^{2}}-4(2)(-5)}}{2(2)}\end{array}[/latex]

Simplify.

[latex]x=\frac{-9\pm \sqrt{121}}{4}=\frac{-9\pm 11}{4}[/latex]

Find the solutions, making sure that the [latex]\pm[/latex] is evaluated for both values.

[latex]\begin{array}{c}x=\frac{-9+11}{4}=\frac{2}{4}=\frac{1}{2}=0.5\\\\\text{or}\\\\x=\frac{-9-11}{4}=\frac{-20}{4}=-5\end{array}[/latex]

Ignore the solution [latex]x=−5[/latex], since the width could not be negative.

Answer

The width of the border should be 0.5 ft.