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Algebra II : Quadratic Roots

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Example Question #1 : Quadratic Roots

Give the solution set of the equation $x ^{2} + 4x -17 = 0$ .

Possible Answers:

$x = -2 \pm i \sqrt{13}$

$x = 2 \pm \sqrt{13}$

$x = -2 \pm i \sqrt{21}$

$x = -2 \pm \sqrt{21}$

$x = 2 \pm i \sqrt{13}$

Correct answer:

$x = -2 \pm \sqrt{21}$

Explanation:

Using the quadratic formula, with a = 1, b = 4, c = -17 :

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

$x = \frac{-(4) \pm \sqrt{(4)^{2}-4 (1)(-17)}}{2 (1)}$

$x = \frac{-4 \pm \sqrt{16-(-68)}}{2 }$

$x = \frac{-4 \pm \sqrt{84}}{2 }$

$x = \frac{-4 \pm \sqrt{4} \cdot \sqrt{21}}{2 }$

$x = \frac{-4 \pm 2 \sqrt{21}}{2 }$

$x = -2 \pm \sqrt{21}$

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Example Question #2 : Quadratic Roots

Give the solution set of the equation $x ^{2} + 4x + 13= 0$ .

Possible Answers:

$x = 2 \pm 3 i$

$x = -2 \pm 3 i$

$x = -3 \pm 2i$

$x = -5 \textup{ or } x= 1$

$x = -1 \textup{ or } x= 5$

Correct answer:

$x = -2 \pm 3 i$

Explanation:

Using the quadratic formula, with a = 1, b = 4, c = 13 :

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

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

$x = \frac{-4 \pm \sqrt{16-52}}{2 }$

$x = \frac{-4 \pm \sqrt{-36}}{2 }$

$x = \frac{-4 \pm i \sqrt{36}}{2 }$

$x = \frac{-4 \pm 6 i }{2 }$

$x = -2 \pm 3 i$

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Example Question #3 : Quadratic Roots

Write a quadratic equation in the form ax^2+bx+c=0 with 2 and -10 as its roots.

Possible Answers:

x^2+10x-2=0

x^2+12x-20=0

x^2+2x-10=0

x^2+8x-10=0

x^2+8x-20=0

Correct answer:

x^2+8x-20=0

Explanation:

Write in the form (x-p)(x-q)=0 where p and q are the roots.

Substitute in the roots:

(x-2)(x-(-10))=0

Simplify:

(x-2)(x+10)=0

Use FOIL and simplify to get

x^2+8x-20=0 .

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Example Question #4 : Quadratic Roots

Find the roots of the following quadratic polynomial:

6x^2+2x-28

Possible Answers:

$-4, \frac{7}{6}$

$2, -\frac{7}{3}$

$\frac{3}{4},-\frac{2}7{}$

This quadratic has no real roots.

$\frac{3}{14},1$

Correct answer:

$2, -\frac{7}{3}$

Explanation:

To find the roots of this equation, we need to find which values of make the polynomial equal zero; we do this by factoring. Factoring is a lot of "guess and check" work, but we can figure some things out. If our binomials are in the form (ax+b)(cx+d) , we know times will be and times will be . With that in mind, we can factor our polynomial to

(2x-4)(3x+7)

To find the roots, we need to find the -values that make each of our binomials equal zero. For the first one it is , and for the second it is $-\frac{7}{3}$ , so our roots are $2, -\frac{7}{3}$ .

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Example Question #5 : Quadratic Roots

Write a quadratic equation in the form $ax^{2}+bx+c=0$ that has and as its roots.

Possible Answers:

$x^{2}+12x+32=0$

$x^{2}+8x+1=0$

$x^{2}-4x-32=0$

$x^{2}+4x-32=0$

$x^{2}-12x+32=0$

Correct answer:

$x^{2}+4x-32=0$

Explanation:

1. Write the equation in the form (x-a)(x-b)=0 where and are the given roots.

(x-4)(x-(-8))=0

(x-4)(x+8)=0

2. Simplify using FOIL method.

$x^{2}+4x-32=0$

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Example Question #6 : Quadratic Roots

Give the solution set of the following equation:

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

Possible Answers:

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

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

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

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

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

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 #7 : Quadratic Roots

Give the solution set of the following equation:

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

Possible Answers:

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

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

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

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

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

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 #8 : Quadratic Roots

Let

(x-3)^2 = 36

Determine the value of x.

Possible Answers:

x=-3, 9

x=6,-6

x=9

x=6

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 #21 : How To Find The Solution To An Equation

Solve for :

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

Possible Answers:

(6,-6)

(3,-9)

(9,3)

None of the other answers

(9,-3)

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 #10 : Quadratic Roots

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

Possible Answers:

x = 14, x = 1

x = 4, x = 5

x = 7, x = 2

x = -7, x = -2

x=0

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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Algebra II : Quadratic Roots

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #1 : Quadratic Roots

Example Question #2 : Quadratic Roots

Example Question #3 : Quadratic Roots

Example Question #4 : Quadratic Roots

Example Question #5 : Quadratic Roots

Example Question #6 : Quadratic Roots

Example Question #7 : Quadratic Roots

Example Question #8 : Quadratic Roots

Example Question #21 : How To Find The Solution To An Equation

Example Question #10 : Quadratic Roots

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