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Algebra II : Algebra II

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Example Questions

Example Question #1441 : Algebra Ii

1-Ax^2=Bx+C

Suppose the equation above has A, B , and such that B^2 = 4(1-C)A . Select which statement must be true about the solutions.

Possible Answers:

There is one real solution and one complex solution.

There are two complex solutions.

There are two real solutions.

There are no solutions.

There is one real solution.

Correct answer:

There is one real solution.

Explanation:

1-Ax^2=Bx+C (1)

B^2 = 4(1-C)A (2)

The first step is conceptualize. What kind of equation is equation (1)? If we rearrange it's clearly a quadratic equation.

1-Ax^2=Bx+C

Ax^2 -Bx+(1-C)=0 (3)

To determine the number of solutions use the discriminate:

______________________________________________________________

Reminder

Recall that for a quadratic equation ax^2+bx+c=0 the general formula for the solution in terms of the constant coefficients is given by:

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

The quantity under the radical is known as the discriminate. If the discriminate is less than zero, there are two complex valued solution. In cases where the discriminate is zero, there is just one real solution $-\frac{b}{2a}$ . If the discriminate is positive, then there are two real solutions.

D = b^2 - 4ca

_____________________________________________________________

The discriminate for equation (3) is written:

D = B^2 -4(1-C)A

We are given that B^2 = 4(1-C)A . Therefore we have:

D = B^2 -4(1-C)A =0

The discriminate is therefore zero, meaning there is only one real solution.

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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 #1441 : Algebra Ii

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

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

Possible Answers:

57, 1 real solution

57, 2 real solutions

-7, 2 real solutions

-7, 2 imaginary solutions

57, 2 imaginary 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 #301 : Intermediate Single Variable Algebra

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

Possible Answers:

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

$x = -2 \pm 3 i$

$x = 2 \pm 3 i$

$x = -3 \pm 2i$

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

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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Algebra II : Algebra II

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #1441 : Algebra Ii

Example Question #141 : Understanding Quadratic Equations

Example Question #1441 : Algebra Ii

Example Question #1 : Quadratic Roots

Example Question #301 : Intermediate Single Variable Algebra

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

All Algebra II Resources

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