Solve H=(4x-3)^frac{2{2}}-2(1-x)(1-x)

$$H = ( 4 x - 3 ) ^ { \frac { 2 } { 2 } } - 2 ( 1 - x ) ( 1 - x )$$

Solve for H

$$H=-2x^{2}+8x-5$$

Solution Steps

Multiply 1-x and 1-x to get \left(1-x\right)^{2}.

$$H=\left(4x-3\right)^{\frac{2}{2}}-2\left(1-x\right)^{2}$$

Divide 2 by 2 to get 1.

$$H=\left(4x-3\right)^{1}-2\left(1-x\right)^{2}$$

Calculate 4x-3 to the power of 1 and get 4x-3.

$$H=4x-3-2\left(1-x\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(1-x\right)^{2}.

$$H=4x-3-2\left(1-2x+x^{2}\right)$$

Use the distributive property to multiply -2 by 1-2x+x^{2}.

$$H=4x-3-2+4x-2x^{2}$$

Subtract 2 from -3 to get -5.

$$H=4x-5+4x-2x^{2}$$

Combine 4x and 4x to get 8x.

$$H=8x-5-2x^{2}$$

Solve for x (complex solution)

$$x=\frac{\sqrt{6-2H}}{2}+2$$

$$x=-\frac{\sqrt{6-2H}}{2}+2$$

Solve for x

$$x=\frac{\sqrt{6-2H}}{2}+2$$

$$x=-\frac{\sqrt{6-2H}}{2}+2\text{, }H\leq 3$$