Solve {left(12-3x/4right)}^2=9/16left(4-xright)^2

$${ \left( \frac{ 12-3x }{ 4 } \right) }^{ 2 } = \frac{ 9 }{ 16 } { \left( 4-x \right) }^{ 2 }$$

Solve for x (complex solution)

$$x\in \mathrm{C}$$

Solution Steps

To raise \frac{12-3x}{4} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{\left(12-3x\right)^{2}}{4^{2}}=\frac{9}{16}\left(4-x\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(4-x\right)^{2}.

$$\frac{\left(12-3x\right)^{2}}{4^{2}}=\frac{9}{16}\left(16-8x+x^{2}\right)$$

Use the distributive property to multiply \frac{9}{16} by 16-8x+x^{2}.

$$\frac{\left(12-3x\right)^{2}}{4^{2}}=9-\frac{9}{2}x+\frac{9}{16}x^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(12-3x\right)^{2}.

$$\frac{144-72x+9x^{2}}{4^{2}}=9-\frac{9}{2}x+\frac{9}{16}x^{2}$$

Calculate 4 to the power of 2 and get 16.

$$\frac{144-72x+9x^{2}}{16}=9-\frac{9}{2}x+\frac{9}{16}x^{2}$$

Divide each term of 144-72x+9x^{2} by 16 to get 9-\frac{9}{2}x+\frac{9}{16}x^{2}.

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

Subtract 9 from both sides.

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

Subtract 9 from 9 to get 0.

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

Add \frac{9}{2}x to both sides.

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

Combine -\frac{9}{2}x and \frac{9}{2}x to get 0.

$$\frac{9}{16}x^{2}=\frac{9}{16}x^{2}$$

Subtract \frac{9}{16}x^{2} from both sides.

$$\frac{9}{16}x^{2}-\frac{9}{16}x^{2}=0$$

Combine \frac{9}{16}x^{2} and -\frac{9}{16}x^{2} to get 0.

$$0=0$$

Compare 0 and 0.

$$\text{true}$$

This is true for any x.

$$x\in \mathrm{C}$$

Solve for x

$$x\in \mathrm{R}$$

Solution Steps

To raise \frac{12-3x}{4} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{\left(12-3x\right)^{2}}{4^{2}}=\frac{9}{16}\left(4-x\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(4-x\right)^{2}.

$$\frac{\left(12-3x\right)^{2}}{4^{2}}=\frac{9}{16}\left(16-8x+x^{2}\right)$$

Use the distributive property to multiply \frac{9}{16} by 16-8x+x^{2}.

$$\frac{\left(12-3x\right)^{2}}{4^{2}}=9-\frac{9}{2}x+\frac{9}{16}x^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(12-3x\right)^{2}.

$$\frac{144-72x+9x^{2}}{4^{2}}=9-\frac{9}{2}x+\frac{9}{16}x^{2}$$

Calculate 4 to the power of 2 and get 16.

$$\frac{144-72x+9x^{2}}{16}=9-\frac{9}{2}x+\frac{9}{16}x^{2}$$

Divide each term of 144-72x+9x^{2} by 16 to get 9-\frac{9}{2}x+\frac{9}{16}x^{2}.

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

Subtract 9 from both sides.

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

Subtract 9 from 9 to get 0.

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

Add \frac{9}{2}x to both sides.

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

Combine -\frac{9}{2}x and \frac{9}{2}x to get 0.

$$\frac{9}{16}x^{2}=\frac{9}{16}x^{2}$$

Subtract \frac{9}{16}x^{2} from both sides.

$$\frac{9}{16}x^{2}-\frac{9}{16}x^{2}=0$$

Combine \frac{9}{16}x^{2} and -\frac{9}{16}x^{2} to get 0.

$$0=0$$

Compare 0 and 0.

$$\text{true}$$

This is true for any x.

$$x\in \mathrm{R}$$