Solve x^2+36=100-32x+4x^2

$${ x }^{ 2 } +36=100-32x+4 { x }^{ 2 }$$

Solve for x

$$x=8$$

$$x = \frac{8}{3} = 2\frac{2}{3} \approx 2.666666667$$

Steps Using Factoring By Grouping

Subtract 100 from both sides.

$$x^{2}+36-100=-32x+4x^{2}$$

Subtract 100 from 36 to get -64.

$$x^{2}-64=-32x+4x^{2}$$

Add 32x to both sides.

$$x^{2}-64+32x=4x^{2}$$

Subtract 4x^{2} from both sides.

$$x^{2}-64+32x-4x^{2}=0$$

Combine x^{2} and -4x^{2} to get -3x^{2}.

$$-3x^{2}-64+32x=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-3x^{2}+32x-64=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -3x^{2}+ax+bx-64. To find a and b, set up a system to be solved.

$$a+b=32 ab=-3\left(-64\right)=192$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 192.

$$1,192 2,96 3,64 4,48 6,32 8,24 12,16$$

Calculate the sum for each pair.

$$1+192=193 2+96=98 3+64=67 4+48=52 6+32=38 8+24=32 12+16=28$$

The solution is the pair that gives sum 32.

$$a=24 b=8$$

Rewrite -3x^{2}+32x-64 as \left(-3x^{2}+24x\right)+\left(8x-64\right).

$$\left(-3x^{2}+24x\right)+\left(8x-64\right)$$

Factor out 3x in the first and -8 in the second group.

$$3x\left(-x+8\right)-8\left(-x+8\right)$$

Factor out common term -x+8 by using distributive property.

$$\left(-x+8\right)\left(3x-8\right)$$

To find equation solutions, solve -x+8=0 and 3x-8=0.

$$x=8 x=\frac{8}{3}$$

Steps Using the Quadratic Formula

Subtract 100 from both sides.

$$x^{2}+36-100=-32x+4x^{2}$$

Subtract 100 from 36 to get -64.

$$x^{2}-64=-32x+4x^{2}$$

Add 32x to both sides.

$$x^{2}-64+32x=4x^{2}$$

Subtract 4x^{2} from both sides.

$$x^{2}-64+32x-4x^{2}=0$$

Combine x^{2} and -4x^{2} to get -3x^{2}.

$$-3x^{2}-64+32x=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$-3x^{2}+32x-64=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -3 for a, 32 for b, and -64 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-32±\sqrt{32^{2}-4\left(-3\right)\left(-64\right)}}{2\left(-3\right)}$$

Square 32.

$$x=\frac{-32±\sqrt{1024-4\left(-3\right)\left(-64\right)}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$x=\frac{-32±\sqrt{1024+12\left(-64\right)}}{2\left(-3\right)}$$

Multiply 12 times -64.

$$x=\frac{-32±\sqrt{1024-768}}{2\left(-3\right)}$$

Add 1024 to -768.

$$x=\frac{-32±\sqrt{256}}{2\left(-3\right)}$$

Take the square root of 256.

$$x=\frac{-32±16}{2\left(-3\right)}$$

Multiply 2 times -3.

$$x=\frac{-32±16}{-6}$$

Now solve the equation x=\frac{-32±16}{-6} when ± is plus. Add -32 to 16.

$$x=-\frac{16}{-6}$$

Reduce the fraction \frac{-16}{-6} to lowest terms by extracting and canceling out 2.

$$x=\frac{8}{3}$$

Now solve the equation x=\frac{-32±16}{-6} when ± is minus. Subtract 16 from -32.

$$x=-\frac{48}{-6}$$

Divide -48 by -6.

$$x=8$$

The equation is now solved.

$$x=\frac{8}{3} x=8$$

Steps for Completing the Square

Add 32x to both sides.

$$x^{2}+36+32x=100+4x^{2}$$

Subtract 4x^{2} from both sides.

$$x^{2}+36+32x-4x^{2}=100$$

Combine x^{2} and -4x^{2} to get -3x^{2}.

$$-3x^{2}+36+32x=100$$

Subtract 36 from both sides.

$$-3x^{2}+32x=100-36$$

Subtract 36 from 100 to get 64.

$$-3x^{2}+32x=64$$

Divide both sides by -3.

$$\frac{-3x^{2}+32x}{-3}=\frac{64}{-3}$$

Dividing by -3 undoes the multiplication by -3.

$$x^{2}+\frac{32}{-3}x=\frac{64}{-3}$$

Divide 32 by -3.

$$x^{2}-\frac{32}{3}x=\frac{64}{-3}$$

Divide 64 by -3.

$$x^{2}-\frac{32}{3}x=-\frac{64}{3}$$

Divide -\frac{32}{3}, the coefficient of the x term, by 2 to get -\frac{16}{3}. Then add the square of -\frac{16}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{32}{3}x+\left(-\frac{16}{3}\right)^{2}=-\frac{64}{3}+\left(-\frac{16}{3}\right)^{2}$$

Square -\frac{16}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{32}{3}x+\frac{256}{9}=-\frac{64}{3}+\frac{256}{9}$$

Add -\frac{64}{3} to \frac{256}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{32}{3}x+\frac{256}{9}=\frac{64}{9}$$

Factor x^{2}-\frac{32}{3}x+\frac{256}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{16}{3}\right)^{2}=\frac{64}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{16}{3}\right)^{2}}=\sqrt{\frac{64}{9}}$$

Simplify.

$$x-\frac{16}{3}=\frac{8}{3} x-\frac{16}{3}=-\frac{8}{3}$$

Add \frac{16}{3} to both sides of the equation.

$$x=8 x=\frac{8}{3}$$