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Algebra II : Finding Roots

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Example Question #61 : Finding Roots

Find the roots of the quadratic equation y=6x^2 -19x + 10

Possible Answers:

$\frac{2}{3} , 0.4$

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

1.5, 2.5

-4, -15

$\frac{2}{3} , 2.5$

Correct answer:

$\frac{2}{3} , 2.5$

Explanation:

This quadratic equation can be solved in several ways, including the quadratic equation or by graphing. It can also be solved by factoring:

For this quadratic, we are looking for 2 numbers that add to -19 and multiply to 6x10=60. The numbers satisfying these conditions are -4 and -15:

6x^2 -4x - 15x + 10 continue factoring

2x(3x - 2 ) - 5 (3x - 2 )

The factors are (2x -5 ) and (3x - 2 ) . Set each equal to zero:

2x-5=0

2x = 5

x=2.5

3x-2 = 0

3x = 2

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

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Example Question #62 : Finding Roots

Find the roots of the quadratic equation y=2x^2 -5x -25

Possible Answers:

-5, 2.5

5, - 2.5

5, 2.5

-10, 5

Correct answer:

5, - 2.5

Explanation:

This quadratic can be solved in several different ways, including by graphing or factoring. You can also use the quadratic formula:

$x= \frac{ 5 \pm \sqrt{ 25 - 4(2)(-25)}}{2 \cdot 2 } = \frac{5 \pm \sqrt{225}}{4} = \frac{5 \pm 15}{4}$

This gives us two answers:

$\frac{5 + 15 } {4} = \frac{20}{4} = 5$

$\frac{5-15}{4} = \frac{-10}{4} = -2.5$

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Example Question #63 : Finding Roots

Given the equation y=4x^2-4x-3 , find a possible root.

Possible Answers:

$-\frac{1}{2}$

$-\frac{1}{4}$

$\frac{2}{3}$

$\frac{1}{8}$

Correct answer:

$-\frac{1}{2}$

Explanation:

Use the quadratic equation to determine roots of a parabolic function.

The quadratic formula is:

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

The equation y=4x^2-4x-3 in the form of y=ax^2+bx+c .

Substitute the known coefficients.

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

Simplify the equation.

$x= \frac{4\pm\sqrt{16+48}}{8} = \frac{4\pm\sqrt{64}}{8} = \frac{4\pm8}{8}$

The fraction can be broken into two.

$\frac{4+8}{8} = \frac{12}{8}=\frac{3}{2}$

$\frac{4-8}{8} = \frac{-4}{8}= -\frac{1}{2}$

Either $-\frac{1}{2}$ or $\frac{3}{2}$ is a possible root.

The answer is: $-\frac{1}{2}$

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Example Question #64 : Finding Roots

Find the roots of this quadratic: 8x^2-24x+18

Possible Answers:

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

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

None of these

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

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

Correct answer:

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

Explanation:

8x^2-24x+18

Simplify the equation by dividing by 2:

4x^2-12x+9

Find the roots using the quadratic equation:

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

$x=\frac{12\pm \sqrt{12^2-4(4)(9)}}{2(4)}\rightarrow \frac{12\pm \sqrt{0}}{8}\rightarrow \boldsymbol{x=\frac{3}{2}}$

*Because the discriminant (part under the radical) comes to 0, we know our quadratic will have just one repeated solution.

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Example Question #65 : Finding Roots

$\frac{20-x}{x}=x$

Possible Answers:

Complex solutions

x=-5

x=4

None of these

x=-5

x=-4

x=4

x=5

Correct answer:

x=-5

x=4

Explanation:

$\frac{20-x}{x}=x$

Multiply both sides by x:

20-x=x^2

Put in standard form:

x^2+x-20

Factor quadratic:

(x+5)(x-4)

$\boldsymbol{x=-5}$

$\boldsymbol{x=4}$

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Example Question #66 : Finding Roots

Find the roots: y=x^2-7x+1

Possible Answers:

$\frac{7\pm 6\sqrt5}{2}$

$\frac{7\pm\sqrt{53}}{2}$

$\frac{7\pm 9\sqrt5}{2}$

$\frac{49\pm 5\sqrt5}{2}$

$\frac{7\pm 3\sqrt5}{2}$

Correct answer:

$\frac{7\pm 3\sqrt5}{2}$

Explanation:

Evaluate by first writing the quadratic equation.

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

The given equation is already in standard form. Write the coefficients.

a=1, b=-7, c=1

$x=\frac{-(-7)\pm\sqrt{(-7)^2-4(1)(1)}}{2(1)} =\frac{7\pm\sqrt{49-4}}{2}=\frac{7\pm\sqrt{45}}{2}$

Simplify the radical using perfect squares.

$\sqrt{45}=\sqrt{9}\times \sqrt5= 3\sqrt5$

The answer is: $\frac{7\pm 3\sqrt5}{2}$

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Example Question #67 : Finding Roots

Find the roots, if any: y=3x^2-10x+1

Possible Answers:

$\frac{5}{3} \pm \frac{i\sqrt{22}}{3}$

$\textup{There are no roots.}$

$\frac{5}{3} \pm \frac{\sqrt{22}}{3}$

$\frac{5}{3} \pm \frac{2\sqrt{7}}{3}$

$\frac{5}{3} \pm \frac{2i\sqrt{7}}{3}$

Correct answer:

$\frac{5}{3} \pm \frac{\sqrt{22}}{3}$

Explanation:

Write the quadratic equation in order to solve for the roots.

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

The coefficients of the parabola y=ax^2+bx+c will be used to substitute as values into the given formula.

a=3,b=-10, c=1

Substitute the values into the equation.

$x=\frac{-(-10)\pm \sqrt{(-10)^2-4(3)(1)}}{2(3)} =\frac{10\pm \sqrt{100-12}}{6}$

Simplify the fraction.

$x=\frac{10\pm \sqrt{100-12}}{6} = \frac{10\pm \sqrt{88}}{6}$

Rewrite $\sqrt{88}$ using factors of perfect squares.

$\sqrt{88} = \sqrt{4}\cdot \sqrt{22} = 2\sqrt{22}$

Replace the term.

$x= \frac{10 \pm 2\sqrt{22} }{6} =\frac{5}{3} \pm \frac{\sqrt{22}}{3}$

The answer is: $\frac{5}{3} \pm \frac{\sqrt{22}}{3}$

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Example Question #68 : Finding Roots

A ball is launched straight up from the ground at time t=0 , and the height of the ball from the ground at time is described by the function h(t)=-x^2 +8x . Find the time at which the ball returns to the ground.

Possible Answers:

t=8

t=6

t=2

t=4

t=10

Correct answer:

t=8

Explanation:

In order to solve this problem, we think about how to translate what is asked of us in the problem into a mathematical notion. The problem asks us to find when the ball returns to the ground. We know that the ball is launched from the ground at time t=0 , and if we plug t=0 into our original function, we will find that at time t=0 , h(0)=(0)^2-8(0)=0 . Therefore, our "ground" is located at h(t)=0 , which is also the horizontal axis. This means that if we are looking for when our ball returns to the ground, mathematically we are looking for values of which make h(t)=0 . In other words, we are looking for the zeros (or horizontal intercepts) of our function.

To find the zeros, we first need to factor the function. The steps of this process can be seen below.

h(t)=-t^2+8t

h(t)=-(t^2-8t)

h(t)=-(t)(t-8)

h(t)=(-t)(t-8)

So, we can see that we have two factors - (-t) and (t-8) . Now, as usual when solving for zeros, we need to set the right side of our function equal to 0 and solve for . Because if either of these factors is zero, the entire function will equal zero (as anything multiplied by zero is zero), we can consider each factor separately when set to zero.

0=-t

0=t

0=t-8

8=t

Thus, we can see that our two zeros are t=0,8 . Since we know that t=0 is when the ball was first launched, we know that when the ball reaches the ground again must be t=8 . We can check this by plugging t=8 back into the original function and ensuring that it evaluates to zero.

h(t) = -t^2+8t

h(8)=-(8)^2+8(8)

h(8) = -(64)+64

h(8)=0

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Example Question #69 : Finding Roots

Evaluate the roots for: y=2x^2-2x+2

Possible Answers:

$\frac{1}{4}\pm\frac{i\sqrt6}{4}$

$\frac{1}{4}\pm\frac{3i\sqrt3}{2}$

$\frac{1}{2}\pm\frac{i\sqrt3}{2}$

$\frac{1}{3}\pm\frac{i\sqrt6}{3}$

$\frac{1}{4}\pm\frac{i\sqrt3}{4}$

Correct answer:

$\frac{1}{2}\pm\frac{i\sqrt3}{2}$

Explanation:

Write the quadratic formula.

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

The coefficients of the variables can be determined by the given equation in standard form: y=ax^2+bx+c .

a=2, b=-2,c=2

Substitute the terms into the equation.

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

Simplify the terms.

$\frac{2\pm\sqrt{4-16}}{4} = \frac{2\pm\sqrt{-12}}{4} = \frac{2\pm 2i\sqrt{3}}{4}$

The answer is: $\frac{1}{2}\pm\frac{i\sqrt3}{2}$

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Example Question #70 : Finding Roots

What is a possible root to the quadratic equation? y=-5x^2-x-1

Possible Answers:

$-\frac{1}{10}- \frac{i\sqrt{21}}{10}$

$-\frac{1}{10}- \frac{i\sqrt{19}}{10}$

$-\frac{1}{2}- \frac{i\sqrt{19}}{2}$

$-\frac{1}{2}- \frac{i\sqrt{19}}{10}$

$\frac{1}{2}- \frac{i\sqrt{19}}{2}$

Correct answer:

$-\frac{1}{10}- \frac{i\sqrt{19}}{10}$

Explanation:

Write the quadratic equation.

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

The given polynomial y=-5x^2-x-1 is already in standard form.

y=ax^2+bx+c

a=-5,b=-1,c =-1

Substitute the values into the quadratic formula.

$x=\frac{-(-1)\pm \sqrt{(-1)^2-4(-5)(-1)}}{2(-5)}$

Simplify the top and bottom.

$x=\frac{1\pm \sqrt{1-20}}{-10} = \frac{1\pm \sqrt{-19}}{-10}$

The roots will be complex since we have a negative term inside the radical.

$\frac{1\pm \sqrt{-19}}{-10} =\frac{1\pm i\sqrt{19}}{-10} =- \frac{1}{10}\pm \frac{i\sqrt{19}}{10}$

A possible root is: $-\frac{1}{10}- \frac{i\sqrt{19}}{10}$

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Algebra II : Finding Roots

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #61 : Finding Roots

Example Question #62 : Finding Roots

Example Question #63 : Finding Roots

Example Question #64 : Finding Roots

Example Question #65 : Finding Roots

Example Question #66 : Finding Roots

Example Question #67 : Finding Roots

Example Question #68 : Finding Roots

Example Question #69 : Finding Roots

Example Question #70 : Finding Roots

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