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Algebra II : Intermediate Single-Variable Algebra

Study concepts, example questions & explanations for Algebra II

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Example Questions

Example Question #172 : Solving Quadratic Equations

Solve x^2+8x-2=0 .

Possible Answers:

$x={-4\pm {3\sqrt{2}}}$

$x={4\pm {3\sqrt{2}}}$

$x={-4\pm {3\sqrt{6}}}$

$x={-4\pm {6\sqrt{2}}}$

$x={-8\pm {3\sqrt{2}}}$

Correct answer:

$x={-4\pm {3\sqrt{2}}}$

Explanation:

x^2+8x-2=0

This particular function cannot be factored therefore, use the quadratic formula to solve.

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

Since the function is in the form ax^2+bx+c where

a=1,b=8, c=-2

the quadratic formula becomes as follows.

$\frac{-8\pm {\sqrt{8^2-4(1)(-2)}}}{2(1)}$

$\frac{-8\pm {\sqrt{64+8}}}{2}$

$\frac{-8\pm {\sqrt{72}}}{2}$

$\frac{-8\pm {\sqrt{9}\times\sqrt{4}\times\sqrt{2}}}{2}$

$\frac{-8\pm {3\times2\times\sqrt{2}}}{2}$

${-4\pm {3\sqrt{2}}}$

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Example Question #1621 : Algebra Ii

Use the quadratic formula to solve for x:

4x+x^2+2

Possible Answers:

$-4\pm\sqrt{8}$

$-4\pm2\sqrt{2}$

$-4\pm\sqrt{2}$

$-2\pm\sqrt{2}$

${-2\pm2\sqrt{2}}$

Correct answer:

$-2\pm\sqrt{2}$

Explanation:

To solve this problem, you must first rewrite the equation into ax^2+bx+c form (quadratic form).

After this you plug the numbers into the following quadratic equation:

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

Which upon doing you get:

$x=\frac{-4\pm \sqrt{16-8}}{2}$

This simplifies to:

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

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Example Question #174 : Solving Quadratic Equations

Use the quadratic formula to find the roots of the quadratic,

2x^2 + 3x-7 = 0

Possible Answers:

$x = \left\{\frac{3}{4} + \frac{\sqrt{65}}{2}, \frac{3}{4} - \frac{\sqrt{65}}{2} \right\}$

$x = \left\{-\frac{3}{4} + \frac{3\sqrt{6}}{4}i, -\frac{3}{4} - \frac{3\sqrt{6}}{4}i \right\}$

$x = \left\{-\frac{3}{4} + \frac{\sqrt{65}}{4}, -\frac{3}{4} - \frac{\sqrt{65}}{4} \right\}$

$x = \left\{-\frac{3}{4} + \frac{3\sqrt{6}}{4} -\frac{3}{4} - \frac{3\sqrt{6}}{4} \right\}$

$x = \left\{-\frac{3}{4} + \frac{\sqrt{65}}{2}i, -\frac{3}{4} - \frac{\sqrt{65}}{2}i \right\}$

Correct answer:

$x = \left\{-\frac{3}{4} + \frac{\sqrt{65}}{4}, -\frac{3}{4} - \frac{\sqrt{65}}{4} \right\}$

Explanation:

2x^2 +3x - 7 = 0

Recall the general form of a quadratic,

ax^2 +bx+c = 0

The solution set has the form,

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

For our particular case, a = 2 , b = 3 , and c = -7

$x = \frac{-3 \pm \sqrt{3^2 - 4(2)(-7)}}{2(2)}$

$x = \frac{-3 \pm \sqrt{65}}{4}$

$x = \left\{-\frac{3}{4} + \frac{\sqrt{65}}{4}, -\frac{3}{4} - \frac{\sqrt{65}}{4} \right\}$

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Example Question #31 : Quadratic Formula

Use the quadratic formula to find the roots of the following equation.

2x^2-12x+16=0

Possible Answers:

x=12, x=8

x=-6,x=-12

x=-12,x=16

x=4,x=2

x=2,x=-4

Correct answer:

x=4,x=2

Explanation:

2x^2-12x+16=0

First, simplify the equation, so that the numbers are easier to work with. We can see that we can factor a 2 from each term.

2(x^2-6x+8)=0

Now we can divide both sides by 2, to further simplify.

$\frac{2}{2}(x^2-6x+8)=\frac{0}{2}$

x^2-6x+8 = 0

Now that we have simplified we can apply the quadratic formula.

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

where a, b, and c are the constants defined as follows:

ax^2+bx+c = 0

This means our a is 1, our b is -6, and our c is 8.

Finally, lets plug the numbers into the formula:

$\frac{-(-6)\pm \sqrt{(-6)^2-4(1)(8)}}{2(1)}$

$\frac{6\pm \sqrt{36-32}}{2}$

$\frac{-6\pm \sqrt{4}}{2}$

$\frac{-6\pm 2}{2}$

$\frac{-4}{2}$ and $\frac{-8}{2}$

or more simply:

2, 4

These are the roots of the equation.

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Example Question #176 : Solving Quadratic Equations

Find the roots of the equation using the quadratic equation.

4x^2-9x+3

Possible Answers:

$x =\frac{9\pm \sqrt{33}}{8}$

x=4,x=-9

$x=\pm \frac{\sqrt{33}}{-8}$

$x=9+\frac{\sqrt{33}}{8},x = 9-\frac{\sqrt{33}}{8}$

$x= \frac{4\pm \sqrt{65}}{2}$

Correct answer:

$x =\frac{9\pm \sqrt{33}}{8}$

Explanation:

4x^2-9x+3

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

where a, b, and c are the constants defined as follows:

ax^2+bx+c = 0

This means our a is 1, our b is -6, and our c is 8.

Finally, lets plug the numbers into the formula:

$\frac{-(-9)\pm \sqrt{(-9)^2-4(4)(3)}}{2(4)}$

$\frac{9\pm \sqrt{81-48}}{8}$

$\frac{9\pm \sqrt{33}}{8}$

These are the roots of the equation.

$x =\frac{9\pm \sqrt{33}}{8}$

Remember that using $\pm$ is the exact same as writing:

$x =\frac{9+ \sqrt{33}}{8}, x =\frac{9- \sqrt{33}}{8}$

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Example Question #177 : Solving Quadratic Equations

Use the quadratic formula to find the answer of the following quadratic equation.

3x^2-9x+2

Possible Answers:

$x=\frac{9+\sqrt{57}}{6},x=\frac{9-\sqrt{57}}{6}$

$x=3+\sqrt{57},x=3-\sqrt{57}$

$x=\frac{3+\sqrt{57}}{2},x=\frac{3-\sqrt{57}}{2}$

$x=\frac{9+\sqrt{59}}{6},x=\frac{9-\sqrt{59}}{6}$

$x=\frac{9+2\sqrt{57}}{6},x=\frac{9-2\sqrt{57}}{6}$

Correct answer:

$x=\frac{9+\sqrt{57}}{6},x=\frac{9-\sqrt{57}}{6}$

Explanation:

The quadratic equation is:

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

$\\a=3 \\b=-9 \\c=2$

Therefore:

$\frac{-(-9)\pm \sqrt{(-9)^2-4(3)(2)}}{2(3)}$

Which gives the answer:

$x=\frac{9\pm \sqrt{57}}{6}$

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Example Question #31 : Quadratic Formula

Solve for x:

3x^2+12x+15=0

Possible Answers:

$x=-2\pm i$

x=-2+i

$x=2\pm i$

$x=-12\pm 6i$

Correct answer:

$x=-2\pm i$

Explanation:

For a quadratic function

ax^2+bx+c

the quadratic formula states that

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

Using the formula for our function, we get

$x=\frac{-12\pm \sqrt{(12)^2-4(3)(15)}}{2(3)}=\frac{-12\pm \sqrt{-36}}{6}$

Notice that we have a negative under the square root. This means that we must use the imaginary number

$i=\sqrt{-1}$

and our roots will be imaginary.

Simplifying using i, we get

$x=\frac{-12\pm 6i}{6}=-2\pm i$

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Example Question #179 : Solving Quadratic Equations

Find the roots using the quadratic formula

x^2+7x+7

Possible Answers:

$-7 \pm \sqrt{21}$

$\frac{-7\pm \sqrt{21}}{2}$

$\frac{7\pm \sqrt{21}}{2}$

$\frac{-7}{2} \pm \sqrt{21}$

$7 \pm \sqrt{21}$

Correct answer:

$\frac{-7\pm \sqrt{21}}{2}$

Explanation:

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

For this problem

a=1, the coefficient on the x^2 term

b=7, the coefficient on the x term

c=7, the constant term

$\frac{-7\pm \sqrt{7^2-4(1)(7)}}{2(1)}\Rightarrow \frac{-7\pm \sqrt{21}}{2}$

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Example Question #180 : Solving Quadratic Equations

Find a root for: y=15x^2+1

Possible Answers:

$i\sqrt{15}$

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

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

$2i\sqrt{15}$

-5i

Correct answer:

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

Explanation:

Write the quadratic equation that applies for y=ax^2+bx+c .

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

There is no term. Substitute the known coefficients from the polynomial.

$x=\frac{\pm \sqrt{-4(15)(1)}}{2(15)}$

Simplify the numerator and denominator.

$x=\frac{ \pm\sqrt{-60}}{30} = \pm\frac{\sqrt{-1} \cdot 2 \sqrt{15}}{30} = \pm \frac{i\sqrt{15}}{15}$

The roots will be imaginary.

One of the possible answers is: $-\frac{i\sqrt{15}}{15}$

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Example Question #331 : Quadratic Equations And Inequalities

Solve: -x^2+2x+2=0

Possible Answers:

$1\pm \sqrt{3}$

None of these

$2\pm \sqrt{3}$

$1\pm \sqrt{2}$

$3\pm \sqrt{2}$

Correct answer:

$1\pm \sqrt{3}$

Explanation:

Solve using the Quadratic formula: $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

-x^2+2x+2=0

Since our quadratic is in standard form ax^2+bx+c , just plug in the values from the equation.

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

Simplify within the radical:

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

Simplify the radical:

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

Divide both numerators by the denominator to simplify:

$\boldsymbol{x=1\pm \sqrt{3}}$

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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 #172 : Solving Quadratic Equations

Example Question #1621 : Algebra Ii

Example Question #174 : Solving Quadratic Equations

Example Question #31 : Quadratic Formula

Example Question #176 : Solving Quadratic Equations

Example Question #177 : Solving Quadratic Equations

Example Question #31 : Quadratic Formula

Example Question #179 : Solving Quadratic Equations

Example Question #180 : Solving Quadratic Equations

Example Question #331 : Quadratic Equations And Inequalities

All Algebra II Resources

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