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GED Math : Quadratic Equations

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Example Questions

Example Question #6 : Solving By Other Methods

Solve the following for x by completing the square:

$x^{2}+8x-3=0$

Possible Answers:

$x=-4+\sqrt{19}$

$x=-4+\sqrt{19}$ or $x=-4-\sqrt{19}$

$x=-4+\sqrt{67}$ or $x=-4-\sqrt{67}$

$x=-4+\sqrt{3}$ or $x=-4-\sqrt{3}$

$x=4+\sqrt{19}$ or $x=4-\sqrt{19}$

Correct answer:

$x=-4+\sqrt{19}$ or $x=-4-\sqrt{19}$

Explanation:

To complete the square, we need to get our variable terms on one side and our constant terms on the other.

1) $x^{2}+8x-3=0$

$x^{2}+8x=3$

2) To make a perfect square trinomial, we need to take one-half of the x-term and square said term. Add the squared term to both sides.

$x^{2}+8x+{\color{Blue} 16}=3+{\color{Blue} 16}$

$x^{2}+8x+16=19$

3) We now have a perfect square trinomial on the left side which can be represented as a binomial squared. We should check to make sure.

* $a^{2}+2ab+b^{2}=(a+b)^{2}$ (standard form)

In our equation:

$(x)^{2}+8x+(4)^{2}$

a=x

b=4

2ab=2(x)(4)=8x (CHECK)

4) Represent the perfect square trinomial as a binomial squared:

$x^{2}+4x+16=(x+4)^{2}$

5) Take the square root of both sides:

$(x+4)^{2}=19$

$\sqrt{(x+4)^{2}}=_{-}^{+}\sqrt{19}$

$(x+4)=_{-}^{+}\sqrt{19}$

6) Solve for x

$x= -4_{-}^{+}\sqrt{19}$

$x=-4+\sqrt{19}$ or $x=-4-\sqrt{19}$

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Example Question #5 : Solving By Other Methods

What are the roots of $6x^{2}+16x-70=0$

Possible Answers:

x=14

x=14 or x=-30

$x=\frac{7}{3}$ or x=-5

$x=\frac{7}{3}$

$x=\frac{-7}{3}$ or x=5

Correct answer:

$x=\frac{7}{3}$ or x=-5

Explanation:

$6x^{2}+16x-70=0$ involves rather large numbers, so the Quadratic Formula is applicable here.

$x=\frac{-b_{-}^{+}\sqrt{b^{2}-4ac}}{2a}$

$x=\frac{-16_{-}^{+}\sqrt{(16)^{2}-4(6)(-70)}}{2(6)}$

$x=\frac{-16_{-}^{+}\sqrt{256+1680}}{12}$

$x=\frac{-16_{-}^{+}\sqrt{1936}}{12}$

$x=\frac{-16_{-}^{+}44}{12}$

$x=\frac{-16+44}{12}$ or $x=\frac{-16-44}{12}$

$x=\frac{28}{12}$ $x=\frac{-60}{12}$

$x=\frac{\textbf{7}}{\textbf{3}}$ $x=\textbf{-5}$

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Example Question #961 : Ged Math

Solve the following by using the Quadratic Formula:

$x^{2}+ 4=0$

Possible Answers:

x=2i

x=2

x=-2

$x=_{-}^{+}\textrm{2}i$

No solution

Correct answer:

$x=_{-}^{+}\textrm{2}i$

Explanation:

The Quadratic Formula:

$x=\frac{-b_{-}^{+}\sqrt{b^{2}-4ac}}{2a}$

Plugging into the Quadratic Formula, we get

$x=\frac{-0_{-}^{+}\sqrt{(0)^{2}-4(1)(4)}}{2(1)}$

$x=\frac{_{-}^{+}\sqrt{0-16}}{2}$

$x=\frac{_{-}^{+}\sqrt{-16}}{2}$

*The square root of a negative number will involve the use of complex numbers

$\sqrt{-16}=\sqrt{(16)(-1)}=(\sqrt{16})(\sqrt{-1})=4(\sqrt{-1})$

$\sqrt{-1}=i$

Therefore, $\sqrt{-16}=4i$

$x=\frac{_{-}^{+}4i}{2}$

$x=_{-}^{+}2i$

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Example Question #92 : Quadratic Equations

A rectangular yard has a width of w and a length two more than three times the width. The area of the yard is 120 square feet. Find the length of the yard.

Possible Answers:

6 feet

20 feet

89 feet

24 feet

5 feet

Correct answer:

20 feet

Explanation:

The area of the garden is 120 square feet. The width is given by w, and the length is 2 more than 3 times the width. Going by the order of operations implied, we have length given by 3w+2.

(length) x (width) = area (for a rectangle)

(3w+2)(w)=120

$3w^{2}+2w=120$

In order to solve for w, we need to set the equation equal to 0.

$3w^{2}+2w-120=0$

To solve this we should use the Quadratic Formula:

$w=\frac{-b_{-}^{+}\sqrt{b^{2}-4ac}}{2a}$

$w=\frac{-2_{-}^{+}\sqrt{(2)^{2}-4(3)(-120)}}{2(3)}$

$w=\frac{-2_{-}^{+}\sqrt{4+1440}}{6}$

$w=\frac{-2_{-}^{+}\sqrt{1444}}{6}$

$w=\frac{-2_{-}^{+}38}{6}$

$w=\frac{-2+38}{6}$ $w=\frac{-2-38}{6}$

$w=\frac{36}{6}$ $w=\frac{-40}{6}$

w=6 $w=-6\frac{2}{3}$ (reject)

The width is 6 feet, so the length is 3(6)+2=20 or 20 feet.

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Example Question #93 : Quadratic Equations

Complete the square to solve for in the equation $x^{2}+3x-13=0$

Possible Answers:

$x= \frac{3_{-}^{+}\sqrt{61}}{2}$

$x= \frac{-3+\sqrt{61}}{2}$

$x=-\frac{3}{2}+\sqrt{13}$

$x= \frac{-3_{-}^{+}\sqrt{61}}{2}$

$x=-\frac{3}{2}+\sqrt{13}$ or $x=-\frac{3}{2}-\sqrt{13}$

Correct answer:

$x= \frac{-3_{-}^{+}\sqrt{61}}{2}$

Explanation:

1) Get all of the variables on one side and the constants on the other.

$x^{2}+3x-13=0$

$x^{2}+3x=13$

2) Get a perfect square trinomial on the left side. One-half the x-term, which will be squared. Add squared term to both sides.

$x^{2}+3x+{\color{Blue}\frac{9}{4}}=13+{\color{Blue} \frac{9}{4}}$

$x^{2}+3x+{\color{Blue} \frac{9}{4}}=\frac{52}{4}+{\color{Blue} \frac{9}{4}}$

$x^{2}+3x+\frac{9}{4}=\frac{61}{4}$

3) We have a perfect square trinomial on the left side

$a^{2}+2ab+b^{2}=(a+b)^{2}$

$x^{2}+3x+\frac{9}{4}=(x+\frac{3}{2})^{2}=\frac{61}{4}$

4) $(x+\frac{3}{2})^{2}=\frac{61}{4}$

5) $\sqrt{(x+\frac{3}{2})^{2}}=_{-}^{+}\sqrt{\frac{61}{4}}$

6) $x+\frac{3}{2}=_{-}^{+}\sqrt{\frac{61}{4}}$

7) $x+\frac{3}{2}=_{-}^{+}\frac{\sqrt{61}}{\sqrt{4}}$

8) $x+\frac{3}{2}=_{-}^{+}\frac{\sqrt{61}}{2}$

9) $x= \frac{-3_{-}^{+}\sqrt{61}}{2}$

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Example Question #12 : Solving By Other Methods

Solve the following quadratic equation for x by completing the square:

$4x^{2}-12x+7=0$

Possible Answers:

$x=\frac{-7}{4}$ or $x=\frac{-1}{4}$

$x=\frac{3+\sqrt{2}}{2}$ or $x=\frac{3-\sqrt{2}}{2}$

$x=\frac{5}{2}$

$x=\frac{-3+\sqrt{2}}{2}$ or $x=\frac{-3-\sqrt{2}}{2}$

$x=\frac{3+\sqrt{2}}{2}$

Correct answer:

$x=\frac{3+\sqrt{2}}{2}$ or $x=\frac{3-\sqrt{2}}{2}$

Explanation:

This quadratic equation needs to be solved by completing the square.

1) Get all of the x-terms on the left side, and the constants on the right side.

$4x^{2}-12x+7=0$

$4x^{2}-12x=-7$

2) To put this equation into terms are more common with completing the square, we can make a coefficient of 1 in front of the $x^{2}$ term.

$\frac{4x^{2}}{4}-\frac{12x}{4}=\frac{-7}{4}$

$x^{2}-3x=\frac{-7}{4}$

3) We need to make the left side of the equation into a "perfect square trinomial." To do this, we take one-half of the coefficient in front of the x, square it, and add it to both sides.

$x^{2}-3x + {\color{Blue} (\frac{-3}{2})^{2}}=\frac{-7}{4}+{\color{Blue} (\frac{-3}{2})^{2}}$

$x^{2}-3x+\frac{9}{4}=\frac{-7}{4}+\frac{9}{4}$

$x^{2}-3x+\frac{9}{4}=\frac{2}{4}=\frac{1}{2}$

The left side is a perfect square trinomial.

4) We can represent a perfect square trinomial as a binomial squared.

$(x-\frac{3}{2})^{2}=\frac{1}{2}$

5) Take the square root of both sides

$\sqrt{(x-\frac{3}{2})^{2}}=_{-}^{+}\sqrt{\frac{1}{2}}$

$x-\frac{3}{2}=_{-}^{+}\sqrt{\frac{1}{2}}$

$x-\frac{3}{2}=_{-}^{+}\frac{\sqrt{2}}{2}$

6) Solve for x

$x=_{-}^{+}\frac{\sqrt{2}}{2}+\frac{3}{2}$

$x=\frac{3_{-}^{+}\sqrt{2}}{2}$

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GED Math : Quadratic Equations

Study concepts, example questions & explanations for GED Math

All GED Math Resources

Example Questions

Example Question #6 : Solving By Other Methods

Example Question #5 : Solving By Other Methods

Example Question #961 : Ged Math

Example Question #92 : Quadratic Equations

Example Question #93 : Quadratic Equations

Example Question #12 : Solving By Other Methods

All GED Math Resources

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