Quadratic Equations - GED Math

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Question

Answer

This is a quadratic equation, but it is not in standard form.

We express it in standard form as follows, using the FOIL technique:

$x \cdot x - x \cdot 3 - 5 \cdot x + 5 \cdot 3 = 24$

$x ^{2}-3 x - 5x+ 15 = 24$

$x ^{2}-8 x + 15 = 24$

$x ^{2}-8 x + 15- 24 = 24 - 24$

$x ^{2}-8 x -9 = 0$

Now factor the quadratic expression on the left. It can be factored as

where .

By trial and error we find that , so

$x ^{2}-8 x -9 = 0$

can be rewritten as

.

Set each linear binomial equal to 0 and solve separately:

$x - 9 = 0 \Rightarrow x = 9$

$x + 1 \Rightarrow x = -1$

The solution set is $\left { -1, 9\right }$ .

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Question

Subtract:

$\left (3.1x^{2}+ 4.5x + 0. 5 \right ) - (1.7x^{2} - 3.9x - 3.2)$

Answer

$\left (3.1x^{2}+ 4.5x + 0. 5 \right ) - (1.7x^{2} - 3.9x - 3.2)$

can be determined by subtracting the coefficients of like terms. We can do this vertically as follows:

$\begin{matrix} ; ; 3.1x^{2}+ 4.5x + 0. 5 \ - \underline{(1.7x^{2} - 3.9x - 3.2) } \end{matrix}$

By switching the symbols in the second expression we can transform this to an addition problem, and add coefficients:

$\begin{matrix} ; ; 3.1x^{2}+ 4.5x + 0. 5 \ \underline{-1.7x^{2} +3.9x +3.2 }\ ; ; 1.4x^{2} + 8.4x +3.7\end{matrix}$

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Question

Add:

$\left (3.1x^{2}+ 4.5x + 0. 5 \right ) + (1.7x^{2} - 3.9x - 3.2)$

Answer

$\left (3.1x^{2}+ 4.5x + 0. 5 \right ) + (1.7x^{2} - 3.9x - 3.2)$

can be determined by adding the coefficients of like terms. We can do this vertically as follows:

$\begin{matrix} ; ; 3.1x^{2}+ 4.5x + 0. 5 \ + \underline{1.7x^{2} - 3.9x - 3.2 }\ ; ; 4.8x^{2} + 0.6x -2.7\end{matrix}$

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Question

Which of the following expressions is equivalent to the product?

$\left ( \frac{1}{5x} - 5x \right )\left ( \frac{1}{5x}+5x\right )$

Answer

Use the difference of squares pattern

$(A + B)(A - B) = A^{2} - B^{2}$

with $A = \frac{1}{5x}$ and :

$\left ( \frac{1}{5x} - 5x \right )\left ( \frac{1}{5x}+5x\right )$

$=\left ( \frac{1}{5x} \right )^{2} - \left ( 5x\right )^{2}$

$= \frac{1}{25x^{2}} - 25x ^{2}$

$= \frac{1}{25x^{2}} - \frac{25x ^{2} \cdot 25x ^{2} }{25x ^{2} }$

$= \frac{1}{25x^{2}} - \frac{625x ^{4} }{25x ^{2} }$

$= \frac{1-625x ^{4} }{25x^{2}}$

$= \frac{-625x ^{4} + 1 }{25x^{2}}$

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Question

Which of the following expressions is equivalent to the product?

$\left ( \frac{1}{3x} - \frac{2}{3}\right )\left ( \frac{1}{3x} + \frac{2}{3}\right )$

Answer

Use the difference of squares pattern

$(A + B)(A - B) = A^{2} - B^{2}$

with $A = \frac{1}{3x}$ and $B = \frac{2}{3}$ :

$\left ( \frac{1}{3x} - \frac{2}{3}\right )\left ( \frac{1}{3x} + \frac{2}{3}\right )$

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

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

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

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

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

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Question

Simplify:

$\frac{x^3+7x^2+10x}{x^2+3x+2}$

Answer

Start by factoring the numerator. Notice that each term in the numerator has an , so we can write the following:

Next, factor the terms in the parentheses. You will want two numbers that multiply to and add to .

Next, factor the denominator. For the denominator, we will want two numbers that multiply to and add to .

Now that both the denominator and numerator have been factored, rewrite the fraction in its factored form.

$\frac{x^3+7x^2+10x}{x^2+3x+2}=\frac{x(x+5)(x+2)}{(x+2)(x+1)}$

Cancel out any terms that appear in both the numerator and denominator.

$\frac{x(x+5)(x+2)}{(x+2)(x+1)}=\frac{x(x+5)}{x+1}$

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Question

Simplify the following expression:

$\frac{x^2+5x+6}{x^2+3x+2}$

Answer

Start by factoring the numerator.

To factor the numerator, you will need to find numbers that add up to and multiply to .

Next, factor the denominator.

To factor the denominator, you will need to find two numbers that add up to and multiply to .

Rewrite the fraction in its factored form.

$\frac{x^2+5x+6}{x^2+3x+2}=\frac{(x+2)(x+3)}{(x+1)(x+2)}$

Since is found in both numerator and denominator, they will cancel out.

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

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Question

Simplify:

$\frac{x^{2}+6x-7}{2x^{2}-8x+6}$

Answer

We need to factor both the numerator and the denominator to determine what can cancel each other out.

If we factor the numerator:

$x^{2}+6x-7$

Two numbers which add to 6 and multiply to give you -7.

Those numbers are 7 and -1.

$x^{2}+6x-7=\textbf{(x+7)(x-1)}$

If we factor the denominator:

$2x^{2}-8x+6$

First factor out a 2 $\rightarrow$ $2(x^{2}-4x+3)$

Two numbers which add to -4 and multiply to give you 3

Those numbers are -3 and -1

$2x^{2}-8x+6=\textbf{2(x-1)(x-3)}$

Now we can re-write our expression with a product of factors:

$\frac{(x+7)(x-1)}{2(x-1)(x-3)}$

We can divide and to give us 1, so we are left with

$\frac{(x+7)}{2(x-3)}$

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Question

Answer

This is a quadratic equation, but it is not in standard form.

We express it in standard form as follows, using the FOIL technique:

$x \cdot x - x \cdot 3 - 5 \cdot x + 5 \cdot 3 = 24$

$x ^{2}-3 x - 5x+ 15 = 24$

$x ^{2}-8 x + 15 = 24$

$x ^{2}-8 x + 15- 24 = 24 - 24$

$x ^{2}-8 x -9 = 0$

Now factor the quadratic expression on the left. It can be factored as

where .

By trial and error we find that , so

$x ^{2}-8 x -9 = 0$

can be rewritten as

.

Set each linear binomial equal to 0 and solve separately:

$x - 9 = 0 \Rightarrow x = 9$

$x + 1 \Rightarrow x = -1$

The solution set is $\left { -1, 9\right }$ .

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Question

Subtract:

$\left (3.1x^{2}+ 4.5x + 0. 5 \right ) - (1.7x^{2} - 3.9x - 3.2)$

Answer

$\left (3.1x^{2}+ 4.5x + 0. 5 \right ) - (1.7x^{2} - 3.9x - 3.2)$

can be determined by subtracting the coefficients of like terms. We can do this vertically as follows:

$\begin{matrix} ; ; 3.1x^{2}+ 4.5x + 0. 5 \ - \underline{(1.7x^{2} - 3.9x - 3.2) } \end{matrix}$

By switching the symbols in the second expression we can transform this to an addition problem, and add coefficients:

$\begin{matrix} ; ; 3.1x^{2}+ 4.5x + 0. 5 \ \underline{-1.7x^{2} +3.9x +3.2 }\ ; ; 1.4x^{2} + 8.4x +3.7\end{matrix}$

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Question

Add:

$\left (3.1x^{2}+ 4.5x + 0. 5 \right ) + (1.7x^{2} - 3.9x - 3.2)$

Answer

$\left (3.1x^{2}+ 4.5x + 0. 5 \right ) + (1.7x^{2} - 3.9x - 3.2)$

can be determined by adding the coefficients of like terms. We can do this vertically as follows:

$\begin{matrix} ; ; 3.1x^{2}+ 4.5x + 0. 5 \ + \underline{1.7x^{2} - 3.9x - 3.2 }\ ; ; 4.8x^{2} + 0.6x -2.7\end{matrix}$

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Question

Which of the following expressions is equivalent to the product?

$\left ( \frac{1}{5x} - 5x \right )\left ( \frac{1}{5x}+5x\right )$

Answer

Use the difference of squares pattern

$(A + B)(A - B) = A^{2} - B^{2}$

with $A = \frac{1}{5x}$ and :

$\left ( \frac{1}{5x} - 5x \right )\left ( \frac{1}{5x}+5x\right )$

$=\left ( \frac{1}{5x} \right )^{2} - \left ( 5x\right )^{2}$

$= \frac{1}{25x^{2}} - 25x ^{2}$

$= \frac{1}{25x^{2}} - \frac{25x ^{2} \cdot 25x ^{2} }{25x ^{2} }$

$= \frac{1}{25x^{2}} - \frac{625x ^{4} }{25x ^{2} }$

$= \frac{1-625x ^{4} }{25x^{2}}$

$= \frac{-625x ^{4} + 1 }{25x^{2}}$

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Question

Which of the following expressions is equivalent to the product?

$\left ( \frac{1}{3x} - \frac{2}{3}\right )\left ( \frac{1}{3x} + \frac{2}{3}\right )$

Answer

Use the difference of squares pattern

$(A + B)(A - B) = A^{2} - B^{2}$

with $A = \frac{1}{3x}$ and $B = \frac{2}{3}$ :

$\left ( \frac{1}{3x} - \frac{2}{3}\right )\left ( \frac{1}{3x} + \frac{2}{3}\right )$

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

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

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

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

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

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Question

Simplify:

$\frac{x^3+7x^2+10x}{x^2+3x+2}$

Answer

Start by factoring the numerator. Notice that each term in the numerator has an , so we can write the following:

Next, factor the terms in the parentheses. You will want two numbers that multiply to and add to .

Next, factor the denominator. For the denominator, we will want two numbers that multiply to and add to .

Now that both the denominator and numerator have been factored, rewrite the fraction in its factored form.

$\frac{x^3+7x^2+10x}{x^2+3x+2}=\frac{x(x+5)(x+2)}{(x+2)(x+1)}$

Cancel out any terms that appear in both the numerator and denominator.

$\frac{x(x+5)(x+2)}{(x+2)(x+1)}=\frac{x(x+5)}{x+1}$

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Question

Simplify the following expression:

$\frac{x^2+5x+6}{x^2+3x+2}$

Answer

Start by factoring the numerator.

To factor the numerator, you will need to find numbers that add up to and multiply to .

Next, factor the denominator.

To factor the denominator, you will need to find two numbers that add up to and multiply to .

Rewrite the fraction in its factored form.

$\frac{x^2+5x+6}{x^2+3x+2}=\frac{(x+2)(x+3)}{(x+1)(x+2)}$

Since is found in both numerator and denominator, they will cancel out.

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

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Question

Simplify:

$\frac{x^{2}+6x-7}{2x^{2}-8x+6}$

Answer

We need to factor both the numerator and the denominator to determine what can cancel each other out.

If we factor the numerator:

$x^{2}+6x-7$

Two numbers which add to 6 and multiply to give you -7.

Those numbers are 7 and -1.

$x^{2}+6x-7=\textbf{(x+7)(x-1)}$

If we factor the denominator:

$2x^{2}-8x+6$

First factor out a 2 $\rightarrow$ $2(x^{2}-4x+3)$

Two numbers which add to -4 and multiply to give you 3

Those numbers are -3 and -1

$2x^{2}-8x+6=\textbf{2(x-1)(x-3)}$

Now we can re-write our expression with a product of factors:

$\frac{(x+7)(x-1)}{2(x-1)(x-3)}$

We can divide and to give us 1, so we are left with

$\frac{(x+7)}{2(x-3)}$

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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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