Solving by Other Methods - GED Math

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Question

Solve for x by using the Quadratic Formula:

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

Answer

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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Question

Solve for by completing the square:

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

Answer

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

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

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

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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Question

Solve for :

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

Answer

$4x ^{2} - 20x + 25$ 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 .

We can rewrite and solve the equation accordingly:

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

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

$x = 5 \div 2$

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

This is the only solution.

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Question

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

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

Answer

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

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

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

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

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.

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

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

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.

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

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

Take the square root of both sides

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

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

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

Solve for x

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

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

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Question

Solve for :

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

Answer

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:

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

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

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

We factor the quadratic expression as

so that and .

By trial and error, we find that

, so the equation becomes

.

Set each linear binomial to 0 and solve separately:

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

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Question

Solve for :

Answer

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:

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

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

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

We factor the quadratic expression as

so that and .

By trial and error, we find that

, so the equation becomes

Set each linear binomial to 0 and solve separately:

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

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Question

Rounded to the nearest tenths place, what is solution to the equation ?

Answer

Solve the equation by using the quadratic formula:

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

For this equation, . Plug these values into the quadratic equation and to solve for .

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

and

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Question

What is the solution to the equation ? Round your answer to the nearest tenths place.

Answer

Recall the quadratic equation:

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

For the given equation, . 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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Question

What is the solution to the equation ? Round your answer to the nearest hundredths place.

Answer

Solve this equation by using the quadratic equation:

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

For the equation ,

Plug it in to the equation to solve for .

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

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

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

and

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Question

Solve the following by using the Quadratic Formula:

$x^{2}+ 4=0$

Answer

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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Question

Solve the following for x by completing the square:

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

Answer

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

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

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

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.

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

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

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:

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

(CHECK)

Represent the perfect square trinomial as a binomial squared:

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

Take the square root of both sides:

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

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

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

Solve for x

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

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

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Question

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.

Answer

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)

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

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:

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

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

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

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

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

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

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

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

The width is 6 feet, so the length is or 20 feet.

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Question

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

Answer

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

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

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

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

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

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

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

We have a perfect square trinomial on the left side

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

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

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

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

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

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

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

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

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Question

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

Answer

$6x^{2}+16x-70=0$ 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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Question

Solve for x by using the Quadratic Formula:

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

Answer

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

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

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

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

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

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

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

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

Compare your answer with the correct one above

Question

Solve for by completing the square:

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

Answer

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

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

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

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:

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

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

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

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

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

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

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

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

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

Compare your answer with the correct one above

Question

Solve for :

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

Answer

$4x ^{2} - 20x + 25$ 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 .

We can rewrite and solve the equation accordingly:

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

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

$x = 5 \div 2$

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

This is the only solution.

Compare your answer with the correct one above

Question

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

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

Answer

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

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

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

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

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.

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

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

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.

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

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

Take the square root of both sides

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

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

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

Solve for x

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

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

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Question

Solve for :

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

Answer

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:

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

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

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

We factor the quadratic expression as

so that and .

By trial and error, we find that

, so the equation becomes

.

Set each linear binomial to 0 and solve separately:

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

Compare your answer with the correct one above

Question

Solve for :

Answer

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:

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

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

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

We factor the quadratic expression as

so that and .

By trial and error, we find that

, so the equation becomes

Set each linear binomial to 0 and solve separately:

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

Compare your answer with the correct one above

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