Solving by Other Methods - GED Math

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Question

Solve for by completing the square:

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

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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 :

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

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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

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

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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.

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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.

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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 for x by using the Quadratic Formula:

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

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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 the following quadratic equation for x by completing the square:

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

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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 the following by using the Quadratic Formula:

$x^{2}+ 4=0$

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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:

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

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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.

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

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

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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 by completing the square:

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

Tap to reveal answer

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 :

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

Tap to reveal answer

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

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

Tap to reveal answer

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.

Tap to reveal answer

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.

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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 for x by using the Quadratic Formula:

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

Tap to reveal answer

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 the following quadratic equation for x by completing the square:

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

Tap to reveal answer

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 the following by using the Quadratic Formula:

$x^{2}+ 4=0$

Tap to reveal answer

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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