Solve 4^2+x^2={left(3/2x-4right)}^2

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

Solve for x

$$x = \frac{48}{5} = 9\frac{3}{5} = 9.6$$

$$x=0$$

Steps Using Factoring

Calculate 4 to the power of 2 and get 16.

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\frac{3}{2}x-4\right)^{2}.

$$16+x^{2}=\frac{9}{4}x^{2}-12x+16$$

Subtract \frac{9}{4}x^{2} from both sides.

$$16+x^{2}-\frac{9}{4}x^{2}=-12x+16$$

Combine x^{2} and -\frac{9}{4}x^{2} to get -\frac{5}{4}x^{2}.

$$16-\frac{5}{4}x^{2}=-12x+16$$

Add 12x to both sides.

$$16-\frac{5}{4}x^{2}+12x=16$$

Subtract 16 from both sides.

$$16-\frac{5}{4}x^{2}+12x-16=0$$

Subtract 16 from 16 to get 0.

$$-\frac{5}{4}x^{2}+12x=0$$

Factor out x.

$$x\left(-\frac{5}{4}x+12\right)=0$$

To find equation solutions, solve x=0 and -\frac{5x}{4}+12=0.

$$x=0 x=\frac{48}{5}$$

Steps Using the Quadratic Formula

Calculate 4 to the power of 2 and get 16.

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\frac{3}{2}x-4\right)^{2}.

$$16+x^{2}=\frac{9}{4}x^{2}-12x+16$$

Subtract \frac{9}{4}x^{2} from both sides.

$$16+x^{2}-\frac{9}{4}x^{2}=-12x+16$$

Combine x^{2} and -\frac{9}{4}x^{2} to get -\frac{5}{4}x^{2}.

$$16-\frac{5}{4}x^{2}=-12x+16$$

Add 12x to both sides.

$$16-\frac{5}{4}x^{2}+12x=16$$

Subtract 16 from both sides.

$$16-\frac{5}{4}x^{2}+12x-16=0$$

Subtract 16 from 16 to get 0.

$$-\frac{5}{4}x^{2}+12x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -\frac{5}{4} for a, 12 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-12±\sqrt{12^{2}}}{2\left(-\frac{5}{4}\right)}$$

Take the square root of 12^{2}.

$$x=\frac{-12±12}{2\left(-\frac{5}{4}\right)}$$

Multiply 2 times -\frac{5}{4}.

$$x=\frac{-12±12}{-\frac{5}{2}}$$

Now solve the equation x=\frac{-12±12}{-\frac{5}{2}} when ± is plus. Add -12 to 12.

$$x=\frac{0}{-\frac{5}{2}}$$

Divide 0 by -\frac{5}{2} by multiplying 0 by the reciprocal of -\frac{5}{2}.

$$x=0$$

Now solve the equation x=\frac{-12±12}{-\frac{5}{2}} when ± is minus. Subtract 12 from -12.

$$x=-\frac{24}{-\frac{5}{2}}$$

Divide -24 by -\frac{5}{2} by multiplying -24 by the reciprocal of -\frac{5}{2}.

$$x=\frac{48}{5}$$

The equation is now solved.

$$x=0 x=\frac{48}{5}$$

Steps for Completing the Square

Calculate 4 to the power of 2 and get 16.

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\frac{3}{2}x-4\right)^{2}.

$$16+x^{2}=\frac{9}{4}x^{2}-12x+16$$

Subtract \frac{9}{4}x^{2} from both sides.

$$16+x^{2}-\frac{9}{4}x^{2}=-12x+16$$

Combine x^{2} and -\frac{9}{4}x^{2} to get -\frac{5}{4}x^{2}.

$$16-\frac{5}{4}x^{2}=-12x+16$$

Add 12x to both sides.

$$16-\frac{5}{4}x^{2}+12x=16$$

Subtract 16 from both sides.

$$-\frac{5}{4}x^{2}+12x=16-16$$

Subtract 16 from 16 to get 0.

$$-\frac{5}{4}x^{2}+12x=0$$

Divide both sides of the equation by -\frac{5}{4}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{-\frac{5}{4}x^{2}+12x}{-\frac{5}{4}}=\frac{0}{-\frac{5}{4}}$$

Dividing by -\frac{5}{4} undoes the multiplication by -\frac{5}{4}.

$$x^{2}+\frac{12}{-\frac{5}{4}}x=\frac{0}{-\frac{5}{4}}$$

Divide 12 by -\frac{5}{4} by multiplying 12 by the reciprocal of -\frac{5}{4}.

$$x^{2}-\frac{48}{5}x=\frac{0}{-\frac{5}{4}}$$

Divide 0 by -\frac{5}{4} by multiplying 0 by the reciprocal of -\frac{5}{4}.

$$x^{2}-\frac{48}{5}x=0$$

Divide -\frac{48}{5}, the coefficient of the x term, by 2 to get -\frac{24}{5}. Then add the square of -\frac{24}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{48}{5}x+\left(-\frac{24}{5}\right)^{2}=\left(-\frac{24}{5}\right)^{2}$$

Square -\frac{24}{5} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{48}{5}x+\frac{576}{25}=\frac{576}{25}$$

Factor x^{2}-\frac{48}{5}x+\frac{576}{25}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{24}{5}\right)^{2}=\frac{576}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{24}{5}\right)^{2}}=\sqrt{\frac{576}{25}}$$

Simplify.

$$x-\frac{24}{5}=\frac{24}{5} x-\frac{24}{5}=-\frac{24}{5}$$

Add \frac{24}{5} to both sides of the equation.

$$x=\frac{48}{5} x=0$$