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Algebra II : Quadratic Equations and Inequalities

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Example Questions

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 #171 : Solving Quadratic Equations

Solve for x:

3x^2+12x+15=0

Possible Answers:

$x=-12\pm 6i$

$x=2\pm i$

$x=-2\pm i$

x=-2+i

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 #1631 : Algebra Ii

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

Possible Answers:

$3\pm \sqrt{2}$

$1\pm \sqrt{3}$

None of these

$1\pm \sqrt{2}$

$2\pm \sqrt{3}$

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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Example Question #331 : Quadratic Equations And Inequalities

Use the quadratic formula to determine a root: y=3x^2-x+3

Possible Answers:

$\frac{1-i\sqrt{37}}{2}$

$\frac{1-i\sqrt{37}}{6}$

$\frac{1-i\sqrt{35}}{2}$

$\frac{1-i\sqrt{37}}{3}$

$\frac{1-i\sqrt{35}}{6}$

Correct answer:

$\frac{1-i\sqrt{35}}{6}$

Explanation:

Write the quadratic formula for polynomials in the form of y=ax^2+bx+c

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

Substitute the known values.

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

Simplify the equation.

$x=\frac{1\pm\sqrt{1-36}}{6} = \frac{1\pm\sqrt{-35}}{6}= \frac{1\pm i\sqrt{35}}{6}$

The roots to this parabola is: $x= \frac{1\pm i\sqrt{35}}{6}$

The answer is $\frac{1-i\sqrt{35}}{6}$ .

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Example Question #41 : Quadratic Formula

Find a root using the quadratic equation: y=3-10x^2

Possible Answers:

$- \frac{\sqrt{30}}{3}$

$- \frac{\sqrt{30}}{10}$

$- \frac{2\sqrt{15}}{3}$

$- \frac{\sqrt{30}}{2}$

$\frac{\sqrt{15}}{3}$

Correct answer:

$- \frac{\sqrt{30}}{10}$

Explanation:

Write the quadratic equation.

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

Rewrite the given equation in y=ax^2+bx+c form.

y=3-10x^2 = -10x^2+0x+3

We can determine the coefficients of the terms.

a=-10

b=0

c=3

Substitute these values into the quadratic equation.

$x=\frac{\pm \sqrt{-4(-10)(3)}}{2(-10)} = \frac{\pm \sqrt{120}}{-20}$

Rewrite this fraction using common factors, and simplify each step.

$\frac{\pm \sqrt{120}}{-20} = \frac{\pm \sqrt{4\times 30}}{-20} = \frac{\pm \sqrt{4}\cdot\sqrt{30}}{-20}=\frac{\pm 2\cdot\sqrt{30}}{-20} =\pm \frac{\sqrt{30}}{10}$

One of the possible answers given is: $- \frac{\sqrt{30}}{10}$

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Example Question #44 : Quadratic Formula

Solve the following equation:

x^4-13x^2+36=0

Possible Answers:

$x=\pm3$

x=-2,2

$x=3,\pm2$

$x=\pm3,\pm2$

$x=\pm3,2$

Correct answer:

$x=\pm3,\pm2$

Explanation:

Let y=x^2 . Then the given equation can be rewritten in terms of as follows:

(x^2)^2-13(x^2)+36=0

$\Rightarrow y^2-13y+36=0$ .

By the quadratic formula,

$y=\frac{13\pm\sqrt{(13^2)-4(1)(36)}}{2(1)}$

$=\frac{13\pm5}{2}$

=9,4

Since y=x^2 , we now have x^2=9 and x^2=4 , which implies that $x=\pm3$ and $x=\pm2$ . Substituting each of these values yields a true statement; hence, the solutions to the original equation are

$x=\pm3,\pm2$ .

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Algebra II : Quadratic Equations and Inequalities

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #31 : Quadratic Formula

Example Question #176 : Solving Quadratic Equations

Example Question #177 : Solving Quadratic Equations

Example Question #171 : Solving Quadratic Equations

Example Question #179 : Solving Quadratic Equations

Example Question #180 : Solving Quadratic Equations

Example Question #1631 : Algebra Ii

Example Question #331 : Quadratic Equations And Inequalities

Example Question #41 : Quadratic Formula

Example Question #44 : Quadratic Formula

All Algebra II Resources

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