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Algebra II : Intermediate Single-Variable Algebra

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Example Questions

Example Question #311 : Intermediate Single Variable Algebra

Give the solution set of the following equation:

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

Possible Answers:

$x=\frac{3\pm i\sqrt{2}}{2}$

$x={-3\pm i\sqrt{3}}$

$x=\frac{-3\pm i\sqrt{3}}{2}$

$x=\frac{3\pm \sqrt{3}}{2}$

$x=\frac{-6\pm i\sqrt{-3}}{2}$

Correct answer:

$x=\frac{-3\pm i\sqrt{3}}{2}$

Explanation:

Use the quadratic formula with a=1 , b=3 and c=3 :

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

$x=\frac{-3\pm \sqrt{3^{2}-4(1)(3)}}{2(1)}$

$x=\frac{-3\pm \sqrt{9-12}}{2}$

$x=\frac{-3\pm \sqrt{-3}}{2}$

$x=\frac{-3\pm i\sqrt{3}}{2}$

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Example Question #1 : Quadratic Roots

Give the solution set of the following equation:

$10x^{2}+4x+5=0$

Possible Answers:

$x=\frac{-4\pm \sqrt {184}}{10}$

$x=\frac{-4\pm i\sqrt {46}}{20}$

$x=\frac{-2\pm \sqrt {46}}{20}$

$x=\frac{-2\pm i\sqrt {46}}{10}$

$x=\frac{-4\pm i\sqrt {46}}{10}$

Correct answer:

$x=\frac{-2\pm i\sqrt {46}}{10}$

Explanation:

Use the quadratic formula with a=10 , b=4 , and c=5 :

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

$x=\frac{-4\pm \sqrt{4^{2}-4(10)(5)}}{2(10)}$

$x=\frac{-4\pm \sqrt{16-200}}{20}$

$x=\frac{-4\pm \sqrt{-184}}{20}$

$x=\frac{-4\pm \sqrt{46\cdot 4\cdot -1}}{20}$

$x=\frac{-4\pm 2i\sqrt{46}}{20}$

$x=\frac{-2\pm i\sqrt{46}}{10}$

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Example Question #144 : Understanding Quadratic Equations

Let

(x-3)^2 = 36

Determine the value of x.

Possible Answers:

x=6,-6

x=6

x=-3, 9

x=9

Correct answer:

x=-3, 9

Explanation:

To solve for x we need to isolate x. We can do this by taking the square root of each side and then doing algebraic operations.

(x-3)^2=36

$\sqrt(x-3)^2=\sqrt36$

$x-3=\pm6$

Now we need to separate our equation in two and solve for each x.

x-3=6 or x-3=-6

x=9 x=-3

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Example Question #151 : Understanding Quadratic Equations

Solve for :

$(x-3)^{2}=36$

Possible Answers:

(9,3)

(9,-3)

None of the other answers

(6,-6)

(3,-9)

Correct answer:

(9,-3)

Explanation:

To solve this equation, you must first eliminate the exponent from the (x-3) by taking the square root of both sides:

$\sqrt{(x-3)^{2}}=\sqrt{36}$

Since the square root of 36 could be either or , there must be 2 values of . So, solve for

x-3=-6

and

x-3=6

to get solutions of (9,-3) .

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Example Question #1 : Quadratic Roots

Find the roots of $y = x^{2} + 9x + 14$ .

Possible Answers:

x = 4, x = 5

x=0

x = 7, x = 2

x = -7, x = -2

x = 14, x = 1

Correct answer:

x = -7, x = -2

Explanation:

When we factor, we are looking for two number that multiply to the constant, , and add to the middle term, . Looking through the factors of , we can find those factors to be and .

Thus, we have the factors:

(x + 2)(x+7) .

To solve for the solutions, set each of these factors equal to zero.

Thus, we get x + 2 = 0 , or x = -2 .

Our second solution is, x + 7 =0 , or x = -7 .

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Example Question #151 : Quadratic Equations And Inequalities

Write a quadratic function in standard form with roots of -1 and 2.

Possible Answers:

f(x)=x^2-2x-2

f(x)=x^2-3x-2

f(x)=x^2-x+2

f(x)=x^2+x-2

f(x)=x^2-x-2

Correct answer:

f(x)=x^2-x-2

Explanation:

From the zeroes we know

f(x) = (x+1)(x-2)

Use FOIL method to obtain:

f(x)=x^2-x-2

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Example Question #312 : Intermediate Single Variable Algebra

Select the quadratic equation that has these roots:

(2x+5)(x-2)

Possible Answers:

2x^2+x-10

2x^2+x+10

2x^2-x+10

None of these.

2x^2-x-10

Correct answer:

2x^2+x-10

Explanation:

(2x+5)(x-2)

FOIL the two factors to find the quadratic equation.

First terms:

$2x\cdot x=2x^2$

Outer terms:

$2x\cdot -2=-4x$

Inner terms:

$5\cdot x=5x$

Last terms:

$5\cdot -2=-10$

Simplify:

2x^2-4x+5x-10

$\boldsymbol{2x^2+x-10}$

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Example Question #313 : Intermediate Single Variable Algebra

Solve for a possible root: y=3x^2-6x+8

Possible Answers:

$1-\frac{\sqrt{25}}{2}$

6+6i

$1-\frac{4i\sqrt{3}}{3}$

$1-\frac{i\sqrt{15}}{3}$

$\textup{There are no possible roots.}$

Correct answer:

$1-\frac{i\sqrt{15}}{3}$

Explanation:

Write the quadratic equation.

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

The equation y=3x^2-6x+8 is in the form y=ax^2+bx+c .

Substitute the proper coefficients into the quadratic equation.

$x= \frac{-(-6)\pm\sqrt{(-6)^2-4(3)(8)}}{2(3)}=\frac{6\pm\sqrt{36-96}}{6}$

$x=\frac{6\pm\sqrt{36-96}}{6} =\frac{6\pm\sqrt{-60}}{6}$

The negative square root can be replaced by the imaginary term . Simplify square root 60 by common factors of numbers with perfect squares.

$\frac{6\pm\sqrt{-60}}{6} = \frac{6\pm i\sqrt{60}}{6} =\frac{6\pm i\sqrt{4} \cdot \sqrt{15}}{6} =\frac{6\pm 2i\sqrt{15}}{6}$

Simplify the fraction.

$\frac{6\pm 2i\sqrt{15}}{6} = 1\pm \frac{i\sqrt{15}}{3}$

A possible root is: $1-\frac{i\sqrt{15}}{3}$

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Example Question #1455 : Algebra Ii

Solve for the roots (if any) of y=-8x^2+28x-12

Possible Answers:

$-\frac{1}{2},3$

$\textup{There are no real roots.}$

$\frac{1}{2},3$

$\frac{1}{2},\frac{5}{2}$

$-\frac{1}{4},5$

Correct answer:

$\frac{1}{2},3$

Explanation:

Pull out a common factor of negative four.

y=-4(2x^2-7x+3)

The term inside the parentheses can be factored.

2x^2-7x+3 = (2x-1)(x-3)

Set the binomials equal to zero and solve for the roots. We can ignore the negative four coefficient.

$2x-1=0, x=\frac{1}{2}$

x-3=0, x=3

The answers are: $\frac{1}{2},3$

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Example Question #1 : Finding Roots

$y=x^{2}-2x-3$

Factor the above function to find the roots of the quadratic equation.

Possible Answers:

x=-3,1

x=-3,-2

x=-1,3

x=2,3

Correct answer:

x=-1,3

Explanation:

Factoring a quadratic equation means doing FOIL backwards. Recall that when you use FOIL, you start with two binomials and end with a trinomial:

$(x+a)(x+b)=x^{2}+(a+b)x+ab$

Now, we're trying to go the other direction -- starting with a trinomial, and going back to two factors.

Here, -3 is equal to , and -2 is equal to a+b . We can use this information to find out what and are, separately. In other words, we have to find two factors of -3 that add up to -2.

Factors of -3:

3*-1 (sum = 2)
-3*1 (sum = -2)

Thus our factored equation should look like this:

y=(x-3)(x+1)

The roots of the quadratic equation are the values of x for which y is 0.

0=(x-3)(x+1)

We know that anything times zero is zero. So the entire expression equals zero when at least one of the factors equals zero.

$x-3=0\: \: \textup{and} \: \: x+1=0$

$x=3 \: \: \textup{and}\: \: -1$

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Algebra II : Intermediate Single-Variable Algebra

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #311 : Intermediate Single Variable Algebra

Example Question #1 : Quadratic Roots

Example Question #144 : Understanding Quadratic Equations

Example Question #151 : Understanding Quadratic Equations

Example Question #1 : Quadratic Roots

Example Question #151 : Quadratic Equations And Inequalities

Example Question #312 : Intermediate Single Variable Algebra

Example Question #313 : Intermediate Single Variable Algebra

Example Question #1455 : Algebra Ii

Example Question #1 : Finding Roots

All Algebra II Resources

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