Quadratic Roots - Algebra 2

Card 0 of 56

Question

Solve for :

Answer

To solve this equation, you must first eliminate the exponent from the by taking the square root of both sides:

Since the square root of 36 could be either or , there must be 2 values of . So, solve for

and

to get solutions of .

Compare your answer with the correct one above

Question

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

Answer

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

Substitute in the roots:

Simplify:

Use FOIL and simplify to get

.

Compare your answer with the correct one above

Question

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

Answer

Using the quadratic formula, with :

$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}$$

Compare your answer with the correct one above

Question

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

Answer

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 }$$

$x = -2 \pm 3 i$

Compare your answer with the correct one above

Question

Let

Determine the value of x.

Answer

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.

$x-3=\pm6$

Now we need to separate our equation in two and solve for each x.

or

Compare your answer with the correct one above

Question

Write a quadratic equation in the form that has and as its roots.

Answer

1. Write the equation in the form where and are the given roots.

2. Simplify using FOIL method.

Compare your answer with the correct one above

Question

Give the solution set of the following equation:

Answer

Use the quadratic formula with , and :

$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}$

Compare your answer with the correct one above

Question

Give the solution set of the following equation:

Answer

Use the quadratic formula with , , and :

$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}$

Compare your answer with the correct one above

Question

Find the roots of the following quadratic polynomial:

Answer

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}$$ , so our roots are $2, -$\frac{7}{3}$$ .

Compare your answer with the correct one above

Question

Find the roots of $y = $x^{2}$ + 9x + 14$ .

Answer

When we factor, we are looking for two number that multiply to the constant, , and add to the middle term, . Looking through the factors of , we can find those factors to be and .

Thus, we have the factors:

.

To solve for the solutions, set each of these factors equal to zero.

Thus, we get , or .

Our second solution is, , or .

Compare your answer with the correct one above

Question

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

Answer

From the zeroes we know

Use FOIL method to obtain:

Compare your answer with the correct one above

Question

Select the quadratic equation that has these roots:

Answer

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:

Compare your answer with the correct one above

Question

Solve for a possible root:

Answer

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}$}{3}$

Compare your answer with the correct one above

Question

Solve for the roots (if any) of

Answer

Pull out a common factor of negative four.

The term inside the parentheses can be factored.

Set the binomials equal to zero and solve for the roots. We can ignore the negative four coefficient.

$2x-1=0, x=\frac{1}{2}$$

The answers are: $$\frac{1}{2}$,3$

Compare your answer with the correct one above

Question

Solve for :

Answer

To solve this equation, you must first eliminate the exponent from the by taking the square root of both sides:

Since the square root of 36 could be either or , there must be 2 values of . So, solve for

and

to get solutions of .

Compare your answer with the correct one above

Question

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

Answer

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

Substitute in the roots:

Simplify:

Use FOIL and simplify to get

.

Compare your answer with the correct one above

Question

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

Answer

Using the quadratic formula, with :

$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}$$

Compare your answer with the correct one above

Question

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

Answer

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 }$$

$x = -2 \pm 3 i$

Compare your answer with the correct one above

Question

Let

Determine the value of x.

Answer

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.

$x-3=\pm6$

Now we need to separate our equation in two and solve for each x.

or

Compare your answer with the correct one above

Question

Write a quadratic equation in the form that has and as its roots.

Answer

1. Write the equation in the form where and are the given roots.

2. Simplify using FOIL method.

Compare your answer with the correct one above

Tap the card to reveal the answer