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Algebra 1 : Algebra 1

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Example Questions

Example Question #1 : Quadratic Equations

Find all of the solutions to the following quadratic equation:

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

Possible Answers:

None of the above

$x = \frac{-7 \pm \sqrt{17}}{4}$

$x = \frac{-7 \pm \sqrt{33}}{4}$

$x = \frac{7 \pm \sqrt{17}}{2}$

$x = \frac{7 \pm \sqrt{17}}{4}$

Correct answer:

$x = \frac{7 \pm \sqrt{17}}{4}$

Explanation:

This requires the use of the quadratic formula. Recall that:

$x = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$ for $ax^{2} + bx + c = 0$ .

For this problem, $a=2,b=-7,\ and\ c=4$ .

So,

$x = \frac{-(-7) \pm \sqrt{(-7)^{2} - 4(2)(4})}{2(2)}$ .

$x = \frac{7 \pm \sqrt{49-32}}{4}$ .

Therefore, the two solutions are:

$x = \frac{7 \pm \sqrt{17}}{4}$

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Example Question #31 : Solving Equations And Inequallities

Solve for .

(x-3)(x+3) =5x+5

Possible Answers:

$-3 \;or\;3$

$-2 \;or\; 7$

No solution

Correct answer:

$-2 \;or\; 7$

Explanation:

Write the equation in standard form by first eliminating parentheses, then moving all terms to the left of the equal sign.

(x-3)(x+3) =5x+5

First: x*x=x^2

Inside: -3*x=-3x

Outside: x*3=3x

Last: -3*3=-9

$x^{2}-3x+3x-9 =5x+5$

x^2-9=5x+5

$x^{2}-5x-14 =0$

Now factor, set each binomial to zero, and solve individually. We are lookig for two numbers with sum and product ; these numbers are -7,2 .

-7*2=-14 and -7+2=-5

(x-7)(x+2) =0

x-7 =0

x=7

x+2 =0

x=-2

The solution set is $\{-2, 7\}$ .

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Example Question #3 : How To Find The Solution To A Quadratic Equation

Solve for : 3x(x+4)=x+20

Possible Answers:

$\left \{ -4, 15\right \}$

$\left \{ -5, \frac{4}{3}\right \}$

$\left \{ -15, 4\right \}$

$\left \{ -20,-4\right \}$

$\left \{ -\frac{4}{3},5\right \}$

Correct answer:

$\left \{ -5, \frac{4}{3}\right \}$

Explanation:

Eliminate parentheses, then write this quadratic equation in standard form, with all nonzero terms on one side:

3x(x+4)=x+20

$3x \cdot x +3x \cdot 4=x+20$

$3x^{2} +12x=x+20$

$3x^{2} +12x-x-20 =0$

$3x^{2} +11x-20 =0$

Now factor the quadratic expression using the -method - split the middle term into two terms whose coefficients add up to 11 and have product $3 \cdot (-20) = -60$ . These numbers are 15,-4

$3x^{2} -4x+15x-20 =0$

x (3x -4) + 5(3x -4)=0

(x + 5)(3x -4) =0

Set each factor to 0:

x+5 = 0

x = -5

3x-4 = 0

3x=4

$x = \frac{4}{3}$

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Example Question #7 : How To Find The Solution To A Quadratic Equation

$Solve\;for\;x,\;(x-3)^2 =9$

Possible Answers:

$\fn_cm 0,6$

6,12

Correct answer:

$\fn_cm 0,6$

Explanation:

$x-3 = 3 \;or \;x-3 = -3$

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Example Question #4 : How To Find The Solution To A Quadratic Equation

$Solve \;for\;p,(p-5)(p+3)=0$

Possible Answers:

5,3

-5,3

5,-3

-5,-3

Correct answer:

5,-3

Explanation:

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Example Question #61 : Systems Of Equations

$Solve \;for\;q,q^2+9q+14=0$

Possible Answers:

-7,2

7,2

7,-2

-7,-2

14,1

Correct answer:

-7,-2

Explanation:

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Example Question #2311 : Algebra 1

$Solve \; for\;t,t^2 + 6t + 8 = 0$

Possible Answers:

t = -4, -2

t= 6, 8

t= 4, 2

t = -1, -8

Correct answer:

t = -4, -2

Explanation:

$Factor \;t^2 + 6t + 8 = 0$

(t+4)(t+2)=0

t=-4,-2

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Example Question #2312 : Algebra 1

$Solve\;for\;x,x^2 + 7x + 15 = 5$

Possible Answers:

x= 3,5

$\fn_cm x= -2, -5$

x= -3,-5

x= 2, 5

Correct answer:

$\fn_cm x= -2, -5$

Explanation:

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Example Question #32 : Solving Equations And Inequallities

Solve for $\small x$ .

$\small x^2+6x+8=0$

Possible Answers:

$\small x=-2\ or\ x=4$

$\small x=2\ or\ x=4$

$\small x=2\ or\ x=-4$

$\small x=-2\ or\ x=-4$

Correct answer:

$\small x=-2\ or\ x=-4$

Explanation:

$\small x^2+6x+8=0$

Solve by factoring. We need to find two factors that multiply to eight and add to six.

$\small 2*4=8\ \text{and}\ 2+4=6$

$\small x^2+6x+8=(x+2)(x+4)=0$

One of these factors must equal zero in order for the equation to be true.

$\small x+2=0\ or\ x+4=0$

$\small x=-2\ or\ x=-4$

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Example Question #2313 : Algebra 1

Solve $3x^{2}-12=0$

Possible Answers:

x=-3, 12

x=3, 2

x=2, -2

x=-3, 2

x=4, -4

Correct answer:

x=2, -2

Explanation:

First, we must factor out any common factors between the two terms. Both 3 and 12 share a factor of 3, so we can "take" 3 out, like this:

$3(x^{2}-4)$ .

Inside the parentheses, it becomes clear that this is a difference of squares problem (a special factor), which can be solved with the equation

$a^{2}-b^{2}=(a-b)(a+b)$ .

Thus, $x^{2}-2^{2}=(x-2)(x+2)$ .

Now, we can set each factor to 0, and solve for :

x-2=0

x=2

and

x+2=0

x=-2 .

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Algebra 1 : Algebra 1

Study concepts, example questions & explanations for Algebra 1

All Algebra 1 Resources

Example Questions

Example Question #1 : Quadratic Equations

Example Question #31 : Solving Equations And Inequallities

Example Question #3 : How To Find The Solution To A Quadratic Equation

Example Question #7 : How To Find The Solution To A Quadratic Equation

Example Question #4 : How To Find The Solution To A Quadratic Equation

Example Question #61 : Systems Of Equations

Example Question #2311 : Algebra 1

Example Question #2312 : Algebra 1

Example Question #32 : Solving Equations And Inequallities

Example Question #2313 : Algebra 1

All Algebra 1 Resources

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