Algebra II - Finding Roots

Evaluate the roots for:

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

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

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

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

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

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: .

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

Find the roots of the function:
$\small f(x)=x^2+4x-12$

$\small \small x=2\ or\ -6$

$\small x=-4\ or\ 3$

$\small x=-3\ or\ 4$

$\small \small x=-2\ or\ 6$

Explanation

Factor:

$\small \small 0=(x-2)(x+6)$

Double check by factoring:

$\small F: (x)(x)$

$\small \small O:(x)(6)=6x$

$\small \small I: (-2)(x)=-2x$

$\small \small L: (-2)(6)=-12$

Add together: $\small x^2+6x-2x-12=x^2+4x-12$

Therefore:

$\small x-2=0\ or x+6=0$

$\small x=2\or-6$

Solve for .

$x=\pm 6, \pm 4$

Explanation

First factor the equation. Find two numbers that multiply to 24 and sum to -10. These numbers are -6 and -4:

Set both expressions equal to 0 and solve for x:

Find the roots of .

$x= \frac{-13}{3} , \frac{13}{3}$

$x= \frac{3}{13}, \frac{-3}{13}$

Explanation

Notice in this question there are only two terms, the exponent value and the constant value. There is also the negative sign between the two. When we look at each term we see that each is a perfect square. Due to the negative sign between the two, this type of quadratic expression can also be written as a difference of squares. We look at the exponential term and see it is

The perfect square factors of this term are and .

Now we look at the constant term

The perfect square factors of this term are and

Now to combine these into the binomial factor form we need to remember it is the difference of perfect squares meaning we will have one subtraction sign and one adding sign, so we get the following:

From here we solve each binomial for x. To do this we set each binomal to zero and solve for x.

$x=\frac{13}{3}$ $x=\frac{-13}{3}$

Find the roots of the following function.

$x=-\frac{3}{2}, -\frac{1}{4}$

$x=-\frac{3}{2}, -\frac{1}{2}$

$x=-\frac{1}{2}, -\frac{1}{4}$

$x=-{3}, -\frac{1}{4}$

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

Explanation

First, set the equation equal to zero.

From here, factor the equation into two binomials.

Now, set each binomial equal to zero and solve for x.

AND

$\frac{3}{-2}=x$ AND $\frac{1}{4}=x$

What are the roots of ?

$x=\frac{1}{2}, 3$

$x=-\frac{1}{2}, 3$

$x=\frac{1}{2}, -3$

$x=\frac{1}{2}, 1$

Explanation

Since this function is already factored and equal to 0, you can just set each expression equal to 0 to get your roots.

$4x-2=0; x=\frac{1}{2}$

and

Which of the following is a possible root of ?

$\frac{3- i\sqrt{151}}{4}$

$-\frac{5}{2}$

$\frac{3- i\sqrt{151}}{2}$

$\frac{3+i\sqrt{71}}{2}$

$\textup{None of the given answers are possible roots.}$

Explanation

Use the quadratic equation to solve for the roots.

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

Substitute the values of the polynomial into the equation.

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

Simplify the quadratic formula.

$x=\frac{+3\pm\sqrt{9-160}}{4} = \frac{3\pm\sqrt{-151}}{4}$

Since we have a negative discriminant, we will have complex roots even though there are no real roots.

The roots are: $x=\frac{3\pm i\sqrt{151}}{4}$

One of the possible roots given is: $\frac{3- i\sqrt{151}}{4}$

Find the roots:

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

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

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

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

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

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.

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

Simplify the radical using perfect squares.

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

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

What are the x-intercepts for

$x=-\frac{1}{4}$

Explanation

The x-intercepts of a quadratic equation are also the solutions. To find them, factor the quadratic equation. After some trial and error, it can be factored to: . Set those expressions equal to to get you x-intercepts. Your answers are: .

Solve for :

$4= 2x^{2} - 7x$

$\left { -\frac{1}{2}, 4 \right }$

$\left { -3, -2\right }$

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

$\left { -\frac{1}{2}, 2 \right }$

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

Explanation

To solve the quadratic equation, $4= 2x^{2} - 7x$ , first set the equation equal to zero and then factor the quadratic to

Since the equation is equal to zero, at least one of the expressions on the left side of the equation must be equal to zero. Therefore, set and solve to get