Algebra II - Quadratic Roots

Write a quadratic equation in the form with 2 and -10 as its roots.

Explanation

Write in the form where p and q are the roots.

Substitute in the roots:

Simplify:

Use FOIL and simplify to get

Select the quadratic equation that has these roots:

None of these.

Explanation

FOIL the two factors to find the quadratic equation.

First terms:

$2x\cdot x=2x^2$

Outer terms:

$2x\cdot -2=-4x$

Inner terms:

$5\cdot x=5x$

Last terms:

$5\cdot -2=-10$

Simplify:

$\boldsymbol{2x^2+x-10}$

Write a quadratic function in standard form with roots of -1 and 2.

Explanation

From the zeroes we know

Use FOIL method to obtain:

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

$x = -2 \pm 3 i$

$x = 2 \pm 3 i$

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

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

$x = -3 \pm 2i$

Explanation

Using the quadratic formula, with :

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

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

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

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

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

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

$x = -2 \pm 3 i$

Solve for a possible root:

$1-\frac{i\sqrt{15}}{3}$

$\textup{There are no possible roots.}$

$1-\frac{\sqrt{25}}{2}$

$1-\frac{4i\sqrt{3}}{3}$

Explanation

Write the quadratic equation.

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

The equation is in the form .

Substitute the proper coefficients into the quadratic equation.

$x= \frac{-(-6)\pm\sqrt{(-6)^2-4(3)(8)}}{2(3)}=\frac{6\pm\sqrt{36-96}}{6}$

$x=\frac{6\pm\sqrt{36-96}}{6} =\frac{6\pm\sqrt{-60}}{6}$

The negative square root can be replaced by the imaginary term . Simplify square root 60 by common factors of numbers with perfect squares.

$\frac{6\pm\sqrt{-60}}{6} = \frac{6\pm i\sqrt{60}}{6} =\frac{6\pm i\sqrt{4} \cdot \sqrt{15}}{6} =\frac{6\pm 2i\sqrt{15}}{6}$

Simplify the fraction.

$\frac{6\pm 2i\sqrt{15}}{6} = 1\pm \frac{i\sqrt{15}}{3}$

A possible root is: $1-\frac{i\sqrt{15}}{3}$

Find the roots of the following quadratic polynomial:

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

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

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

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

This quadratic has no real roots.

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 , we know times will be and times will be . With that in mind, we can factor our polynomial to

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