Solving by Other Methods - GED Math

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Question

Solve for x by using the Quadratic Formula:

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Answer

We have our quadratic equation in the form

The quadratic formula is given as:

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

Using

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

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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

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:

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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.

To put this equation into terms are more common with completing the square, we can make a coefficient of 1 in front of the term.

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.

The left side is a perfect square trinomial.

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

Take the square root of both sides

$x-$\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

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 .

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.

and

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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 the following by using the Quadratic Formula:

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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:

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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.

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.

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.

* (standard form)

In our equation:

(CHECK)

Represent the perfect square trinomial as a binomial squared:

Take the square root of both sides:

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

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)

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

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Answer

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

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

We have a perfect square trinomial on the left side

5)

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

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Answer

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 :

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Answer

When solving a quadratic equation, it is necessary to write it in standard form first - that is, in the form . 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

Solve for :

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Answer

When solving a quadratic equation, it is necessary to write it in standard form first - that is, in the form . This equation is not in this form, so we must get it in this form as follows:

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 x by using the Quadratic Formula:

Tap to reveal answer

Answer

We have our quadratic equation in the form

The quadratic formula is given as:

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

Using

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

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

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:

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.

To put this equation into terms are more common with completing the square, we can make a coefficient of 1 in front of the term.

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.

The left side is a perfect square trinomial.

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

Take the square root of both sides

$x-$\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

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 .

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.

and

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