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High School Math : Intermediate Single-Variable Algebra

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Example Questions

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Example Question #71 : Intermediate Single Variable Algebra

Solve $x^{2}+6x+5=0$

Possible Answers:

$x=2\; or\; x=3$

$x=3\; or\; x=-1$

$x=-5\; or\; x=-1$

$x=-3\; or\; x=-2$

$x=-5\; or\; x=-2$

Correct answer:

$x=-5\; or\; x=-1$

Explanation:

Factor the problem and set each factor equal to zero.

$x^{2}+6x+5=0$ becomes (x+5)(x+1)=0 so $x=-5 \; or\; x=-1$

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Example Question #12 : Solving Quadratic Equations

Solve $x^{2}-8x+12=0$ .

Possible Answers:

$x=6\; or\; x=-8$

$x=-2\; or\; x=2$

$x=6\; or\; x=-4$

$x=-2\; or\; x=2$

$x=6\; or\; x=2$

$x=6\; or\; x=-8$

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

Correct answer:

$x=6\; or\; x=2$

Explanation:

Factor the quadratic equation and set each factor equal to zero:

$x^{2}-8x+12=0$ becomes (x-6)(x-2)=0 so the correct answer is $x=6\; or \; x=2$ .

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Example Question #72 : Intermediate Single Variable Algebra

What are the roots of (x+3)(x-4) ?

Possible Answers:

(3,4)(4,3)

(0,3)(0,4)

(-3,0)(4,0)

(3,0)(-4,0)

(0,-3)(0,-4)

Correct answer:

(-3,0)(4,0)

Explanation:

To find the roots, we need to find the values that would make x=0 . Since there are two parts to (x+3)(x-4) , we will have two roots: one where x+3=0 , and one where x-4=0 .

Solve each one individually:

x+3=0

x=-3

x-4=0

x=4

Therefore, our roots will be (-3,0)(4,0) .

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Example Question #321 : Algebra Ii

What are the roots of (x+5)(x-1) ?

Possible Answers:

$(\frac{1}{5},0)$

(-5,0)(1,0)

(0,5)(0,-1)

(0,-5)(0,1)

(5,0)(-1,0)

Correct answer:

(-5,0)(1,0)

Explanation:

To find the roots, we need to find what would make x=0 . Since there are two parts to (x+5)(x-1) , we will have two roots: one where x+5=0 , and one where x-1=0 .

Solve each individually.

x+5=0

x=-5

x-1=0

x=1

Our two roots will be (-5,0)(1,0) .

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Example Question #75 : Intermediate Single Variable Algebra

What are the roots of (x+7)(x-5) ?

Possible Answers:

(0,7)(0,-5)

$(\frac{5}{7},0)(\frac{7}{5},0)$

(0,-7)(0,5)

(-7,0)(5,0)

(7,0)(-5,0)

Correct answer:

(-7,0)(5,0)

Explanation:

To find the roots, we need to find what would make x=0 . Since there are two parts to (x+7)(x-5) , we will have two roots: one where x+7=0 , and one where x-5=0 .

Solve each individually.

x+7=0

x=-7

x-5=0

x=5

Therefore, our two roots will be at (-7,0)(5,0) .

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Example Question #76 : Intermediate Single Variable Algebra

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

Possible Answers:

$x=2\; \; or\; \; x=6$

$x=-3\; \; or\; \; x=4$

No solutions

$x=3\; \; or\; \; x=4$

$x=2\; \; or\; \; x=-6$

Correct answer:

$x=3\; \; or\; \; x=4$

Explanation:

Factor the equation by looking for two factors that multiply to +12 and add to .

The factors are and , so the equation to solve becomes (x-3)(x-4)=0 .

Next, set each factor equal to zero and solve:

x-3=0 or x-4=0

The solution is x=3 or x=4 .

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Example Question #77 : Intermediate Single Variable Algebra

Solve 2x^2-3x+1 .

Possible Answers:

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

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

x=-1, 1

x=10, 1

x=1

Correct answer:

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

Explanation:

To find the roots of this equation, you can factor it to

(2x-1)(x-1)

Set each of those expressions equal to zero and then solve for . The roots are and $\frac{1}{2}$ .

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Example Question #78 : Intermediate Single Variable Algebra

Find the root(s) of the following quadratic polynomial.

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

Possible Answers:

x =- 2

x=0

x=2

x=2, x=-2

Correct answer:

x =- 2

Explanation:

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

We set the function equal to 0 and factor the equation. By FOIL, we can confirm that (x+2)(x+2) = 0 is equivalent to the given function. Thus, the only zero comes from x + 2 = 0 , and x = -2 . Thus, x = -2 is the only root.

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Example Question #79 : Intermediate Single Variable Algebra

$\textup{What are the possible values of }x \textup{ when }\; \; x^{2}-2x-15=0\:\textup{?}$

Possible Answers:

3,5

-2, -15

-5. 3

-3, 5

3, 5 -5, -3

Correct answer:

-3, 5

Explanation:

$\textup{Factor to solve. The product of the constants in the two roots is -15, while }$

$\textup{their sum is -2. }\:\:\left ( x+3\right )\left ( x-5\right )=0\;\;\;\;x=-3 \textup{ or }x=5$

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Example Question #80 : Intermediate Single Variable Algebra

Solve the quadratic equation using any method:

bx^2+cx+a=0

Possible Answers:

$x=\frac{-c \pm \sqrt{c^2-4ab}}{b}$

$x=\frac{-c \pm \sqrt{c^2-4ab}}{2b}$

$x=\frac{-c \pm \sqrt{4c^2-4ab}}{2b}$

$x=\frac{-2c \pm \sqrt{c^2-4ab}}{2b}$

$x=\frac{-c \pm \sqrt{c^2-2ab}}{2b}$

Correct answer:

$x=\frac{-c \pm \sqrt{c^2-4ab}}{2b}$

Explanation:

Use the quadratic formula to solve:

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

a = b

b = c

c = a

$x = \frac{-(c) \pm \sqrt{(c)^2-4(a)(b)}}{2(b)}$

$x = \frac{-c \pm \sqrt{c^2-4ab}}{2b}$

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High School Math : Intermediate Single-Variable Algebra

Study concepts, example questions & explanations for High School Math

All High School Math Resources

Example Questions

Example Question #71 : Intermediate Single Variable Algebra

Example Question #12 : Solving Quadratic Equations

Example Question #72 : Intermediate Single Variable Algebra

Example Question #321 : Algebra Ii

Example Question #75 : Intermediate Single Variable Algebra

Example Question #76 : Intermediate Single Variable Algebra

Example Question #77 : Intermediate Single Variable Algebra

Example Question #78 : Intermediate Single Variable Algebra

Example Question #79 : Intermediate Single Variable Algebra

Example Question #80 : Intermediate Single Variable Algebra

All High School Math Resources

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