Solve 4^2=x^2+(7/2-x)^2

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

Solve for x

$$x = \frac{\sqrt{79} + 7}{4} \approx 3.972048604$$

$$x=\frac{7-\sqrt{79}}{4}\approx -0.472048604$$

Steps Using the Quadratic Formula

Calculate 4 to the power of 2 and get 16.

$$16=x^{2}+\left(\frac{7}{2}-x\right)^{2}$$

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

$$16=x^{2}+\frac{49}{4}-7x+x^{2}$$

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

$$16=2x^{2}+\frac{49}{4}-7x$$

Swap sides so that all variable terms are on the left hand side.

$$2x^{2}+\frac{49}{4}-7x=16$$

Subtract 16 from both sides.

$$2x^{2}+\frac{49}{4}-7x-16=0$$

Subtract 16 from \frac{49}{4} to get -\frac{15}{4}.

$$2x^{2}-\frac{15}{4}-7x=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$2x^{2}-7x-\frac{15}{4}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -7 for b, and -\frac{15}{4} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-7\right)±\sqrt{\left(-7\right)^{2}-4\times 2\left(-\frac{15}{4}\right)}}{2\times 2}$$

Square -7.

$$x=\frac{-\left(-7\right)±\sqrt{49-4\times 2\left(-\frac{15}{4}\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-7\right)±\sqrt{49-8\left(-\frac{15}{4}\right)}}{2\times 2}$$

Multiply -8 times -\frac{15}{4}.

$$x=\frac{-\left(-7\right)±\sqrt{49+30}}{2\times 2}$$

Add 49 to 30.

$$x=\frac{-\left(-7\right)±\sqrt{79}}{2\times 2}$$

The opposite of -7 is 7.

$$x=\frac{7±\sqrt{79}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{7±\sqrt{79}}{4}$$

Now solve the equation x=\frac{7±\sqrt{79}}{4} when ± is plus. Add 7 to \sqrt{79}.

$$x=\frac{\sqrt{79}+7}{4}$$

Now solve the equation x=\frac{7±\sqrt{79}}{4} when ± is minus. Subtract \sqrt{79} from 7.

$$x=\frac{7-\sqrt{79}}{4}$$

The equation is now solved.

$$x=\frac{\sqrt{79}+7}{4} x=\frac{7-\sqrt{79}}{4}$$

Steps for Completing the Square

Calculate 4 to the power of 2 and get 16.

$$16=x^{2}+\left(\frac{7}{2}-x\right)^{2}$$

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

$$16=x^{2}+\frac{49}{4}-7x+x^{2}$$

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

$$16=2x^{2}+\frac{49}{4}-7x$$

Swap sides so that all variable terms are on the left hand side.

$$2x^{2}+\frac{49}{4}-7x=16$$

Subtract \frac{49}{4} from both sides.

$$2x^{2}-7x=16-\frac{49}{4}$$

Subtract \frac{49}{4} from 16 to get \frac{15}{4}.

$$2x^{2}-7x=\frac{15}{4}$$

Divide both sides by 2.

$$\frac{2x^{2}-7x}{2}=\frac{\frac{15}{4}}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}-\frac{7}{2}x=\frac{\frac{15}{4}}{2}$$

Divide \frac{15}{4} by 2.

$$x^{2}-\frac{7}{2}x=\frac{15}{8}$$

Divide -\frac{7}{2}, the coefficient of the x term, by 2 to get -\frac{7}{4}. Then add the square of -\frac{7}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{7}{2}x+\left(-\frac{7}{4}\right)^{2}=\frac{15}{8}+\left(-\frac{7}{4}\right)^{2}$$

Square -\frac{7}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{7}{2}x+\frac{49}{16}=\frac{15}{8}+\frac{49}{16}$$

Add \frac{15}{8} to \frac{49}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{7}{2}x+\frac{49}{16}=\frac{79}{16}$$

Factor x^{2}-\frac{7}{2}x+\frac{49}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{7}{4}\right)^{2}=\frac{79}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{7}{4}\right)^{2}}=\sqrt{\frac{79}{16}}$$

Simplify.

$$x-\frac{7}{4}=\frac{\sqrt{79}}{4} x-\frac{7}{4}=-\frac{\sqrt{79}}{4}$$

Add \frac{7}{4} to both sides of the equation.

$$x=\frac{\sqrt{79}+7}{4} x=\frac{7-\sqrt{79}}{4}$$