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GED Math : Solving by Other Methods

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Example Questions

Example Question #1 : Solving By Other Methods

Solve for by completing the square:

$\small 4x^2+12x+2=0$

Possible Answers:

$\small x=\frac{-9\pm \sqrt7}{2}$

$\small x=\frac{-3\pm \sqrt7}{2}$

$\small x=\frac{-3\pm \sqrt{-11}}{2}$

$\small x=\frac{-9\pm \sqrt{-11}}{2}$

Correct answer:

$\small x=\frac{-3\pm \sqrt7}{2}$

Explanation:

$\small 4x^2+12x+2=0$

$\small 4x^2+12+2-2=0-2$

$\small 4x^2+12x=-2$

To complete the square, we have to add a number that makes the left side of the equation a perfect square. Perfect squares have the formula $\small \small \small a^2+2ab+b^2$ .

In this case, $\small \small a=2x,\ 2ab=12x$ .

$\small \small 12=2(2x)(b)=4xb; b=3$

$\small b^2=3^2=9$

Add this to both sides:

$\small 4x^2+12x+9=-2+9$

$\small \small 4x^2+12x+9=7$

$\small (2x+3)^2=7$

$\small \sqrt{(2x+3)^2}=\pm \sqrt7$

$\small 2x+3=\pm \sqrt{7}$

$\small 2x+3-3=-3\pm \sqrt7$

$\small 2x=-3\pm \sqrt7$

$\small \frac{2x}{2}=\frac{-3\pm \sqrt7}{2}$

$\small x=\frac{-3\pm \sqrt7}{2}$

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Example Question #2 : Solving By Other Methods

Solve for :

$4x ^{2} - 20x + 25 = 0$

Possible Answers:

$x = 2\frac{1}{2}$

$x = -\frac{2}{5} \textrm{ or }x = \frac{2}{5}$

$x = - 2\frac{1}{2} \textrm{ or }x = 2\frac{1}{2}$

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

Correct answer:

$x = 2\frac{1}{2}$

Explanation:

$4x ^{2} - 20x + 25$ can be demonstrated to be a perfect square polynomial as follows:

$4x ^{2} - 20x + 25 = \left ( 2x\right )^{2} - 2 (2x)(5) + 5^{2}$

It can therefore be factored using the pattern

$A^{2}-2AB + B^{2} = \left (A - B \right )^{2}$

with A = 2x, B = 5 .

We can rewrite and solve the equation accordingly:

$4x ^{2} - 20x + 25 = 0$

$(2x-5) ^{2} = 0$

2x-5 = 0

2x = 5

$x = 5 \div 2$

$x = 2\frac{1}{2}$

This is the only solution.

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Example Question #3 : Solving By Other Methods

Solve for :

x (x - 7) = 30

Possible Answers:

x = 9 or x = 13

x = 2 or x = 6

x = -3 or x = 10

x = 30 or x = 37

Correct answer:

x = -3 or x = 10

Explanation:

When solving a quadratic equation, it is necessary to write it in standard form first - that is, in the form $ax^{2} + bx + c = 0$ . This equation is not in this form, so we must get it in this form as follows:

x (x - 7) = 30

$x \cdot x -x \cdot 7 = 30$

$x ^{2} -7x = 30$

$x ^{2} -7x - 30 = 0$

We factor the quadratic expression as

(x+M)(x+N)

so that MN = -30 and M + N = -7 .

By trial and error, we find that

M = -10, N = 3 , so the equation becomes

(x-10)(x+3) = 0

Set each linear binomial to 0 and solve separately:

x - 10 = 0

x = 10

x + 3 = 0

x = -3

The solution set is $\left \{ -3, 10 \right \}$ .

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Example Question #4 : Solving By Other Methods

Solve for :

$x^{2} - 12 = 6x+4$

Possible Answers:

x = 4

x= -8 or x= 2

x = -4 or x = 4

x= -2 or x= 8

Correct answer:

x= -2 or x= 8

Explanation:

$x^{2} - 12 = 6x+4$

$x^{2} - 12- 6x - 4 = 6x+4 - 6x - 4$

$x^{2} - 6x - 16 = 0$

We factor the quadratic expression as

(x+M)(x+N)

so that MN = -16 and M + N = -6 .

By trial and error, we find that

M = -8, N = 2 , so the equation becomes

(x-8)(x+2) = 0 .

Set each linear binomial to 0 and solve separately:

x-8=0

x= 8

x+2= 0

x= -2

The solutions set is $\left \{ -2, 8\right \}$

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Example Question #4 : Solving By Other Methods

Rounded to the nearest tenths place, what is solution to the equation y=x^2+5x-13 ?

Possible Answers:

x=-6.9

2.9, -4.9

5.1, -2.1

1.9, -6.9

4.4, -1.2

Correct answer:

1.9, -6.9

Explanation:

Solve the equation by using the quadratic formula:

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

For this equation, a=1, b=5, c=-13 . Plug these values into the quadratic equation and to solve for .

$x=\frac{-5\pm\sqrt{5^2-4(1)(-13)}}{2(1)}=\frac{-5\pm\sqrt{77}}{2}$

x=1.9 and -6.9

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Example Question #5 : Solving By Other Methods

What is the solution to the equation x^2+5x-7=0 ? Round your answer to the nearest tenths place.

Possible Answers:

1.1, -6.1

1.1

4.5, -2.1

-6.1

Correct answer:

1.1, -6.1

Explanation:

Recall the quadratic equation:

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

For the given equation, a=1, b=5, c=-7 . Plug these into the equation and solve.

$x=\frac{-5+\sqrt{5^2-4(1)(-7))}}{2(1)}=\frac{-5+\sqrt{25+28}}{2}=\frac{-5+\sqrt{53}}{2}=1.1$

and

$x=\frac{-5-\sqrt{5^2-4(1)(-7))}}{2(1)}=\frac{-5-\sqrt{25+28}}{2}=\frac{-5-\sqrt{53}}{2}=-6.1$

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Example Question #6 : Solving By Other Methods

What is the solution to the equation x^2+4x-10=0 ? Round your answer to the nearest hundredths place.

Possible Answers:

5.74

-5.74, 1.74

-1.74, 5.74

1.74

Correct answer:

-5.74, 1.74

Explanation:

Solve this equation by using the quadratic equation:

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

For the equation x^2+4x-10=0 , a=1, b=4, c=-10

Plug it in to the equation to solve for .

$x=\frac{-4\pm \sqrt{4^2-4(1)(-10)}}{2}$

$x=\frac{-4\pm \sqrt{16+40}}{2}$

$x=\frac{-4\pm \sqrt{56}}{2}$

x=1.74 and x=-5.74

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Example Question #84 : Quadratic Equations

Solve for x by using the Quadratic Formula:

$2x^{2}+7x-85=0$

Possible Answers:

x = -8.5

x = 5

x = 5 or x= -8.5

x = 10 or x = -17

x = -5 or x = 8.5

Correct answer:

x = 5 or x= -8.5

Explanation:

We have our quadratic equation in the form $ax^{2}+bx+c=0$

The quadratic formula is given as:

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

Using $2x^{2}+7x-85=0$

$x=\frac{-7_{-}^{+}{\sqrt{(7)^{2}-4(2)(-85)}}}{2(2)}$

$x=\frac{-7_{-}^{+}{\sqrt{49+680}}}{4}$

$x=\frac{-7_{-}^{+}{\sqrt{729}}}{4}$

$x=\frac{-7_{-}^{+}{27}}{4}$

$x=\frac{-7+27}{4}$ $x=\frac{-7-27}{4}$

$x=\frac{20}{4}$ $x=\frac{-34}{4}$

$\textbf{x=5}$ $\textbf{x=-8.5}$

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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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GED Math : Solving by Other Methods

Study concepts, example questions & explanations for GED Math

All GED Math Resources

Example Questions

Example Question #1 : Solving By Other Methods

Example Question #2 : Solving By Other Methods

Example Question #3 : Solving By Other Methods

Example Question #4 : Solving By Other Methods

Example Question #4 : Solving By Other Methods

Example Question #5 : Solving By Other Methods

Example Question #6 : Solving By Other Methods

Example Question #84 : Quadratic Equations

Example Question #6 : Solving By Other Methods

Example Question #5 : Solving By Other Methods

All GED Math Resources

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