Solve {left(3/2xright)}^2+left(2xright)^2=5/2x^2

$${ \left( \frac{ 3 }{ 2 } x \right) }^{ 2 } + { \left(2x \right) }^{ 2 } = \frac{ 5 }{ 2 } { x }^{ 2 }$$

Solve for x

$$x=0$$

Steps by Finding Square Root

Expand \left(\frac{3}{2}x\right)^{2}.

$$\left(\frac{3}{2}\right)^{2}x^{2}+\left(2x\right)^{2}=\frac{5}{2}x^{2}$$

Calculate \frac{3}{2} to the power of 2 and get \frac{9}{4}.

$$\frac{9}{4}x^{2}+\left(2x\right)^{2}=\frac{5}{2}x^{2}$$

Expand \left(2x\right)^{2}.

$$\frac{9}{4}x^{2}+2^{2}x^{2}=\frac{5}{2}x^{2}$$

Calculate 2 to the power of 2 and get 4.

$$\frac{9}{4}x^{2}+4x^{2}=\frac{5}{2}x^{2}$$

Combine \frac{9}{4}x^{2} and 4x^{2} to get \frac{25}{4}x^{2}.

$$\frac{25}{4}x^{2}=\frac{5}{2}x^{2}$$

Subtract \frac{5}{2}x^{2} from both sides.

$$\frac{25}{4}x^{2}-\frac{5}{2}x^{2}=0$$

Combine \frac{25}{4}x^{2} and -\frac{5}{2}x^{2} to get \frac{15}{4}x^{2}.

$$\frac{15}{4}x^{2}=0$$

Multiply both sides by \frac{4}{15}, the reciprocal of \frac{15}{4}. Anything times zero gives zero.

$$x^{2}=0$$

Take the square root of both sides of the equation.

$$x=0 x=0$$

The equation is now solved. Solutions are the same.

$$x=0$$

Steps Using the Quadratic Formula

Expand \left(\frac{3}{2}x\right)^{2}.

$$\left(\frac{3}{2}\right)^{2}x^{2}+\left(2x\right)^{2}=\frac{5}{2}x^{2}$$

Calculate \frac{3}{2} to the power of 2 and get \frac{9}{4}.

$$\frac{9}{4}x^{2}+\left(2x\right)^{2}=\frac{5}{2}x^{2}$$

Expand \left(2x\right)^{2}.

$$\frac{9}{4}x^{2}+2^{2}x^{2}=\frac{5}{2}x^{2}$$

Calculate 2 to the power of 2 and get 4.

$$\frac{9}{4}x^{2}+4x^{2}=\frac{5}{2}x^{2}$$

Combine \frac{9}{4}x^{2} and 4x^{2} to get \frac{25}{4}x^{2}.

$$\frac{25}{4}x^{2}=\frac{5}{2}x^{2}$$

Subtract \frac{5}{2}x^{2} from both sides.

$$\frac{25}{4}x^{2}-\frac{5}{2}x^{2}=0$$

Combine \frac{25}{4}x^{2} and -\frac{5}{2}x^{2} to get \frac{15}{4}x^{2}.

$$\frac{15}{4}x^{2}=0$$

Multiply both sides by \frac{4}{15}, the reciprocal of \frac{15}{4}. Anything times zero gives zero.

$$x^{2}=0$$

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

$$x=\frac{0±\sqrt{0^{2}}}{2}$$

Take the square root of 0^{2}.

$$x=\frac{0±0}{2}$$

Divide 0 by 2.

$$x=0$$