Solve (x/4)^2:(7-x/4)^2=1:2

$$( \frac { x } { 4 } ) ^ { 2 } : ( \frac { 7 - x } { 4 } ) ^ { 2 } = 1 : 2$$

Solve for x

$$x=7\sqrt{2}-7\approx 2.899494937$$

$$x=-7\sqrt{2}-7\approx -16.899494937$$

Steps Using the Quadratic Formula

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

$$\frac{\frac{x^{2}}{4^{2}}}{\left(\frac{7-x}{4}\right)^{2}}=\frac{1}{2}$$

To raise \frac{7-x}{4} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{\frac{x^{2}}{4^{2}}}{\frac{\left(7-x\right)^{2}}{4^{2}}}=\frac{1}{2}$$

Divide \frac{x^{2}}{4^{2}} by \frac{\left(7-x\right)^{2}}{4^{2}} by multiplying \frac{x^{2}}{4^{2}} by the reciprocal of \frac{\left(7-x\right)^{2}}{4^{2}}.

$$\frac{x^{2}\times 4^{2}}{4^{2}\left(7-x\right)^{2}}=\frac{1}{2}$$

Cancel out 4^{2} in both numerator and denominator.

$$\frac{x^{2}}{\left(-x+7\right)^{2}}=\frac{1}{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(-x+7\right)^{2}.

$$\frac{x^{2}}{x^{2}-14x+49}=\frac{1}{2}$$

Subtract \frac{1}{2} from both sides.

$$\frac{x^{2}}{x^{2}-14x+49}-\frac{1}{2}=0$$

Factor x^{2}-14x+49.

$$\frac{x^{2}}{\left(x-7\right)^{2}}-\frac{1}{2}=0$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of \left(x-7\right)^{2} and 2 is 2\left(x-7\right)^{2}. Multiply \frac{x^{2}}{\left(x-7\right)^{2}} times \frac{2}{2}. Multiply \frac{1}{2} times \frac{\left(x-7\right)^{2}}{\left(x-7\right)^{2}}.

$$\frac{2x^{2}}{2\left(x-7\right)^{2}}-\frac{\left(x-7\right)^{2}}{2\left(x-7\right)^{2}}=0$$

Since \frac{2x^{2}}{2\left(x-7\right)^{2}} and \frac{\left(x-7\right)^{2}}{2\left(x-7\right)^{2}} have the same denominator, subtract them by subtracting their numerators.

$$\frac{2x^{2}-\left(x-7\right)^{2}}{2\left(x-7\right)^{2}}=0$$

Do the multiplications in 2x^{2}-\left(x-7\right)^{2}.

$$\frac{2x^{2}-x^{2}+14x-49}{2\left(x-7\right)^{2}}=0$$

Combine like terms in 2x^{2}-x^{2}+14x-49.

$$\frac{x^{2}+14x-49}{2\left(x-7\right)^{2}}=0$$

Variable x cannot be equal to 7 since division by zero is not defined. Multiply both sides of the equation by 2\left(x-7\right)^{2}.

$$x^{2}+14x-49=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 14 for b, and -49 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-14±\sqrt{14^{2}-4\left(-49\right)}}{2}$$

Square 14.

$$x=\frac{-14±\sqrt{196-4\left(-49\right)}}{2}$$

Multiply -4 times -49.

$$x=\frac{-14±\sqrt{196+196}}{2}$$

Add 196 to 196.

$$x=\frac{-14±\sqrt{392}}{2}$$

Take the square root of 392.

$$x=\frac{-14±14\sqrt{2}}{2}$$

Now solve the equation x=\frac{-14±14\sqrt{2}}{2} when ± is plus. Add -14 to 14\sqrt{2}.

$$x=\frac{14\sqrt{2}-14}{2}$$

Divide -14+14\sqrt{2} by 2.

$$x=7\sqrt{2}-7$$

Now solve the equation x=\frac{-14±14\sqrt{2}}{2} when ± is minus. Subtract 14\sqrt{2} from -14.

$$x=\frac{-14\sqrt{2}-14}{2}$$

Divide -14-14\sqrt{2} by 2.

$$x=-7\sqrt{2}-7$$

The equation is now solved.

$$x=7\sqrt{2}-7 x=-7\sqrt{2}-7$$

Steps for Completing the Square

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

$$\frac{\frac{x^{2}}{4^{2}}}{\left(\frac{7-x}{4}\right)^{2}}=\frac{1}{2}$$

To raise \frac{7-x}{4} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{\frac{x^{2}}{4^{2}}}{\frac{\left(7-x\right)^{2}}{4^{2}}}=\frac{1}{2}$$

Divide \frac{x^{2}}{4^{2}} by \frac{\left(7-x\right)^{2}}{4^{2}} by multiplying \frac{x^{2}}{4^{2}} by the reciprocal of \frac{\left(7-x\right)^{2}}{4^{2}}.

$$\frac{x^{2}\times 4^{2}}{4^{2}\left(7-x\right)^{2}}=\frac{1}{2}$$

Cancel out 4^{2} in both numerator and denominator.

$$\frac{x^{2}}{\left(-x+7\right)^{2}}=\frac{1}{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(-x+7\right)^{2}.

$$\frac{x^{2}}{x^{2}-14x+49}=\frac{1}{2}$$

Variable x cannot be equal to 7 since division by zero is not defined. Multiply both sides of the equation by 2\left(x-7\right)^{2}, the least common multiple of x^{2}-14x+49,2.

$$2x^{2}=\left(x-7\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-7\right)^{2}.

$$2x^{2}=x^{2}-14x+49$$

Subtract x^{2} from both sides.

$$2x^{2}-x^{2}=-14x+49$$

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

$$x^{2}=-14x+49$$

Add 14x to both sides.

$$x^{2}+14x=49$$

Divide 14, the coefficient of the x term, by 2 to get 7. Then add the square of 7 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+14x+7^{2}=49+7^{2}$$

Square 7.

$$x^{2}+14x+49=49+49$$

Add 49 to 49.

$$x^{2}+14x+49=98$$

Factor x^{2}+14x+49. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+7\right)^{2}=98$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+7\right)^{2}}=\sqrt{98}$$

Simplify.

$$x+7=7\sqrt{2} x+7=-7\sqrt{2}$$

Subtract 7 from both sides of the equation.

$$x=7\sqrt{2}-7 x=-7\sqrt{2}-7$$