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Algebra II : Understanding Quadratic Equations

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Example Questions

Example Question #141 : Understanding Quadratic Equations

Given -9x+5+3x^2=0 , what is the value of the discriminant?

Possible Answers:

$^{\sqrt{21}}$

Correct answer:

Explanation:

The correct answer is . The discriminant is equal to b^2-4ac portion of the quadratic formula. In this case, "" corresponds to the coefficient of 3x^2 , "" corresponds to the coefficient of -9x , and "" corresponds to . So, the answer is (-9)^2-4(3)(5) , which is equal to .

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Example Question #141 : Understanding Quadratic Equations

Find the value of the discriminant and state the number of real and imaginary solutions.

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

Possible Answers:

57, 2 imaginary solutions

57, 1 real solution

57, 2 real solutions

-7, 2 imaginary solutions

-7, 2 real solutions

Correct answer:

57, 2 real solutions

Explanation:

Given the quadratic equation of $ax^{2}+bx+c=0$

a=2, b=5, c=-4

The formula for the discriminant is $b^{2}-4ac$ (remember this as a part of the quadratic formula?)

Plugging in values to the discriminant equation:

$5^{2}-4(2)(-4)$

25+32=57

So the discriminant is 57. What does that mean for our solutions? Since it is a positive number, we know that we will have 2 real solutions. So the answer is:

57, 2 real solutions

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Example Question #1 : Quadratic Roots

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

Possible Answers:

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

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

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

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

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

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 #1441 : Algebra Ii

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

Possible Answers:

$x = -2 \pm 3 i$

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

$x = 2 \pm 3 i$

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

$x = -3 \pm 2i$

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 #2 : 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+2x-10=0

x^2+8x-10=0

x^2+10x-2=0

x^2+12x-20=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 #143 : Understanding Quadratic Equations

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

Possible Answers:

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

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

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

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

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

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 #2 : Quadratic Roots

Give the solution set of the following equation:

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

Possible Answers:

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

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

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

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

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

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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Algebra II : Understanding Quadratic Equations

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #141 : Understanding Quadratic Equations

Example Question #141 : Understanding Quadratic Equations

Example Question #1 : Quadratic Roots

Example Question #1441 : Algebra Ii

Example Question #2 : Quadratic Roots

Example Question #4 : Quadratic Roots

Example Question #143 : Understanding Quadratic Equations

Example Question #2 : Quadratic Roots

Example Question #1 : Quadratic Roots

Example Question #144 : Understanding Quadratic Equations

All Algebra II Resources

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