Solve x^2+8{left(sqrt{x}right)}^2-33=0

$${ x }^{ 2 } +8 { \left( \sqrt{ x } \right) }^{ 2 } -33 = 0$$

Solve for x (complex solution)

$$x=-11$$

$$x=3$$

Steps Using Factoring

Calculate \sqrt{x} to the power of 2 and get x.

$$x^{2}+8x-33=0$$

To solve the equation, factor x^{2}+8x-33 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). To find a and b, set up a system to be solved.

$$a+b=8 ab=-33$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -33.

$$-1,33 -3,11$$

Calculate the sum for each pair.

$$-1+33=32 -3+11=8$$

The solution is the pair that gives sum 8.

$$a=-3 b=11$$

Rewrite factored expression \left(x+a\right)\left(x+b\right) using the obtained values.

$$\left(x-3\right)\left(x+11\right)$$

To find equation solutions, solve x-3=0 and x+11=0.

$$x=3 x=-11$$

Steps Using Factoring By Grouping

Calculate \sqrt{x} to the power of 2 and get x.

$$x^{2}+8x-33=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-33. To find a and b, set up a system to be solved.

$$a+b=8 ab=1\left(-33\right)=-33$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -33.

$$-1,33 -3,11$$

Calculate the sum for each pair.

$$-1+33=32 -3+11=8$$

The solution is the pair that gives sum 8.

$$a=-3 b=11$$

Rewrite x^{2}+8x-33 as \left(x^{2}-3x\right)+\left(11x-33\right).

$$\left(x^{2}-3x\right)+\left(11x-33\right)$$

Factor out x in the first and 11 in the second group.

$$x\left(x-3\right)+11\left(x-3\right)$$

Factor out common term x-3 by using distributive property.

$$\left(x-3\right)\left(x+11\right)$$

To find equation solutions, solve x-3=0 and x+11=0.

$$x=3 x=-11$$

Steps Using the Quadratic Formula

Calculate \sqrt{x} to the power of 2 and get x.

$$x^{2}+8x-33=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 8 for b, and -33 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-8±\sqrt{8^{2}-4\left(-33\right)}}{2}$$

Square 8.

$$x=\frac{-8±\sqrt{64-4\left(-33\right)}}{2}$$

Multiply -4 times -33.

$$x=\frac{-8±\sqrt{64+132}}{2}$$

Add 64 to 132.

$$x=\frac{-8±\sqrt{196}}{2}$$

Take the square root of 196.

$$x=\frac{-8±14}{2}$$

Now solve the equation x=\frac{-8±14}{2} when ± is plus. Add -8 to 14.

$$x=\frac{6}{2}$$

Divide 6 by 2.

$$x=3$$

Now solve the equation x=\frac{-8±14}{2} when ± is minus. Subtract 14 from -8.

$$x=-\frac{22}{2}$$

Divide -22 by 2.

$$x=-11$$

The equation is now solved.

$$x=3 x=-11$$

Steps for Completing the Square

Calculate \sqrt{x} to the power of 2 and get x.

$$x^{2}+8x-33=0$$

Add 33 to both sides. Anything plus zero gives itself.

$$x^{2}+8x=33$$

Divide 8, the coefficient of the x term, by 2 to get 4. Then add the square of 4 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+8x+4^{2}=33+4^{2}$$

Square 4.

$$x^{2}+8x+16=33+16$$

Add 33 to 16.

$$x^{2}+8x+16=49$$

Factor x^{2}+8x+16. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+4\right)^{2}=49$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+4\right)^{2}}=\sqrt{49}$$

Simplify.

$$x+4=7 x+4=-7$$

Subtract 4 from both sides of the equation.

$$x=3 x=-11$$

Solve for x

$$x=3$$

Steps Using Factoring

Calculate \sqrt{x} to the power of 2 and get x.

$$x^{2}+8x-33=0$$

To solve the equation, factor x^{2}+8x-33 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). To find a and b, set up a system to be solved.

$$a+b=8 ab=-33$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -33.

$$-1,33 -3,11$$

Calculate the sum for each pair.

$$-1+33=32 -3+11=8$$

The solution is the pair that gives sum 8.

$$a=-3 b=11$$

Rewrite factored expression \left(x+a\right)\left(x+b\right) using the obtained values.

$$\left(x-3\right)\left(x+11\right)$$

To find equation solutions, solve x-3=0 and x+11=0.

$$x=3 x=-11$$

Steps Using Factoring By Grouping

Calculate \sqrt{x} to the power of 2 and get x.

$$x^{2}+8x-33=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-33. To find a and b, set up a system to be solved.

$$a+b=8 ab=1\left(-33\right)=-33$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -33.

$$-1,33 -3,11$$

Calculate the sum for each pair.

$$-1+33=32 -3+11=8$$

The solution is the pair that gives sum 8.

$$a=-3 b=11$$

Rewrite x^{2}+8x-33 as \left(x^{2}-3x\right)+\left(11x-33\right).

$$\left(x^{2}-3x\right)+\left(11x-33\right)$$

Factor out x in the first and 11 in the second group.

$$x\left(x-3\right)+11\left(x-3\right)$$

Factor out common term x-3 by using distributive property.

$$\left(x-3\right)\left(x+11\right)$$

To find equation solutions, solve x-3=0 and x+11=0.

$$x=3 x=-11$$

Steps Using the Quadratic Formula

Calculate \sqrt{x} to the power of 2 and get x.

$$x^{2}+8x-33=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 8 for b, and -33 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-8±\sqrt{8^{2}-4\left(-33\right)}}{2}$$

Square 8.

$$x=\frac{-8±\sqrt{64-4\left(-33\right)}}{2}$$

Multiply -4 times -33.

$$x=\frac{-8±\sqrt{64+132}}{2}$$

Add 64 to 132.

$$x=\frac{-8±\sqrt{196}}{2}$$

Take the square root of 196.

$$x=\frac{-8±14}{2}$$

Now solve the equation x=\frac{-8±14}{2} when ± is plus. Add -8 to 14.

$$x=\frac{6}{2}$$

Divide 6 by 2.

$$x=3$$

Now solve the equation x=\frac{-8±14}{2} when ± is minus. Subtract 14 from -8.

$$x=-\frac{22}{2}$$

Divide -22 by 2.

$$x=-11$$

The equation is now solved.

$$x=3 x=-11$$

Steps for Completing the Square

Calculate \sqrt{x} to the power of 2 and get x.

$$x^{2}+8x-33=0$$

Add 33 to both sides. Anything plus zero gives itself.

$$x^{2}+8x=33$$

Divide 8, the coefficient of the x term, by 2 to get 4. Then add the square of 4 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+8x+4^{2}=33+4^{2}$$

Square 4.

$$x^{2}+8x+16=33+16$$

Add 33 to 16.

$$x^{2}+8x+16=49$$

Factor x^{2}+8x+16. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+4\right)^{2}=49$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+4\right)^{2}}=\sqrt{49}$$

Simplify.

$$x+4=7 x+4=-7$$

Subtract 4 from both sides of the equation.

$$x=3 x=-11$$