Solving Quadratic Equations

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Math › Solving Quadratic Equations

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1

Solve using the quadratic formula:

$a=\frac{5 \pm i\sqrt{7}}{4}$

$a=\frac{6 \pm i\sqrt{7}}{4}$

$a=\frac{4 \pm i\sqrt{7}}{4}$

$a=\frac{3 \pm i\sqrt{7}}{4}$

$a=\frac{7 \pm i\sqrt{7}}{4}$

Explanation

Use the quadratic formula to solve:

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

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

$a = \frac{5 \pm \sqrt{-7}}{4}$

$a = \frac{5 \pm i\sqrt{7}}{4}$

2

Find the sum of the solutions to:

$\frac{-6}{x^2} + \frac{1}{x} + 1 = 0$

Explanation

Multiply both sides of the equation by , to get

This can be factored into the form

So we must solve

and

to get the solutions.

The solutions are:

and their sum is .

3

Complete the square:

$x=\frac{-3 \pm i\sqrt{7}}{4}$

$x=\frac{-2 \pm i\sqrt{7}}{4}$

$x=\frac{-1 \pm i\sqrt{7}}{4}$

$x=\frac{-3 \pm i\sqrt{7}}{3}$

$x=\frac{-2 \pm i\sqrt{7}}{3}$

Explanation

Begin by dividing the equation by and subtracting from each side:

$x^2+\frac{3}{2}x = -1$

Square the value in front of the and add to each side:

$x^2+\frac{3}{2}x +\frac{9}{16} = -1+\frac{9}{16}$

Factor the left side of the equation:

$(x+\frac{3}{4})^2 = \frac{-7}{16}$

Take the square root of both sides and simplify:

$x+\frac{3}{4} = \frac{\pm \sqrt{-7}}{4}$

$x = \frac{-3 \pm i\sqrt{7}}{4}$

4

Solve using the quadratic formula:

$a=\frac{5 \pm i\sqrt{7}}{4}$

$a=\frac{6 \pm i\sqrt{7}}{4}$

$a=\frac{4 \pm i\sqrt{7}}{4}$

$a=\frac{3 \pm i\sqrt{7}}{4}$

$a=\frac{7 \pm i\sqrt{7}}{4}$

Explanation

Use the quadratic formula to solve:

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

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

$a = \frac{5 \pm \sqrt{-7}}{4}$

$a = \frac{5 \pm i\sqrt{7}}{4}$

5

Solve the following quadratic inequality:

$y\leq x^2+4x+3$

$x\leq -3, x\geq -1$

$x\leq -2, x\geq -1$

$x\leq -4, x\geq -1$

$x\leq -3, x\geq -2$

$x\leq -4, x\geq -2$

Explanation

Factor and solve. Since the sign is less than or equal to, we know the inequality will be OR, not AND.

$y\leq x^2+4x+3$

$y\leq (x+3)(x+1)$

$x\leq-3$ or $x\geq -1$

6

Complete the square:

$x=\frac{-3 \pm i\sqrt{7}}{4}$

$x=\frac{-2 \pm i\sqrt{7}}{4}$

$x=\frac{-1 \pm i\sqrt{7}}{4}$

$x=\frac{-3 \pm i\sqrt{7}}{3}$

$x=\frac{-2 \pm i\sqrt{7}}{3}$

Explanation

Begin by dividing the equation by and subtracting from each side:

$x^2+\frac{3}{2}x = -1$

Square the value in front of the and add to each side:

$x^2+\frac{3}{2}x +\frac{9}{16} = -1+\frac{9}{16}$

Factor the left side of the equation:

$(x+\frac{3}{4})^2 = \frac{-7}{16}$

Take the square root of both sides and simplify:

$x+\frac{3}{4} = \frac{\pm \sqrt{-7}}{4}$

$x = \frac{-3 \pm i\sqrt{7}}{4}$

7

Find the sum of the solutions to:

$\frac{-6}{x^2} + \frac{1}{x} + 1 = 0$

Explanation

Multiply both sides of the equation by , to get

This can be factored into the form

So we must solve

and

to get the solutions.

The solutions are:

and their sum is .

8

Solve the following quadratic inequality:

$y\leq x^2+4x+3$

$x\leq -3, x\geq -1$

$x\leq -2, x\geq -1$

$x\leq -4, x\geq -1$

$x\leq -3, x\geq -2$

$x\leq -4, x\geq -2$

Explanation

Factor and solve. Since the sign is less than or equal to, we know the inequality will be OR, not AND.

$y\leq x^2+4x+3$

$y\leq (x+3)(x+1)$

$x\leq-3$ or $x\geq -1$

9

Solve using the quadratic formula:

$x=\frac{-b \pm \sqrt{b^2-12ab}}{2a}$

$x=\frac{b \pm \sqrt{b^2-12ab}}{2a}$

$x=\frac{-b \pm \sqrt{b^2-12ab}}{a}$

$x=\frac{b \pm \sqrt{b^2-12ab}}{a}$

$x=\frac{-b \pm \sqrt{b^2-8ab}}{2a}$

Explanation

Use the quadratic formula to solve:

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

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

$x = \frac{-b \pm \sqrt{b^2-12ab}}{2a}$

10

Solve using the quadratic formula:

$x=\frac{-b \pm \sqrt{b^2-12ab}}{2a}$

$x=\frac{b \pm \sqrt{b^2-12ab}}{2a}$

$x=\frac{-b \pm \sqrt{b^2-12ab}}{a}$

$x=\frac{b \pm \sqrt{b^2-12ab}}{a}$

$x=\frac{-b \pm \sqrt{b^2-8ab}}{2a}$

Explanation

Use the quadratic formula to solve:

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

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

$x = \frac{-b \pm \sqrt{b^2-12ab}}{2a}$

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