Solve (3+x)^2+(4-x)^2=25

$$( 3 + x ) ^ { 2 } + ( 4 - x ) ^ { 2 } = 25$$

Solve for x

$$x=1$$

$$x=0$$

Steps Using Factoring

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

$$9+6x+x^{2}+\left(4-x\right)^{2}=25$$

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

$$9+6x+x^{2}+16-8x+x^{2}=25$$

Add 9 and 16 to get 25.

$$25+6x+x^{2}-8x+x^{2}=25$$

Combine 6x and -8x to get -2x.

$$25-2x+x^{2}+x^{2}=25$$

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

$$25-2x+2x^{2}=25$$

Subtract 25 from both sides.

$$25-2x+2x^{2}-25=0$$

Subtract 25 from 25 to get 0.

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

Factor out x.

$$x\left(-2+2x\right)=0$$

To find equation solutions, solve x=0 and -2+2x=0.

$$x=0 x=1$$

Steps Using the Quadratic Formula

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

$$9+6x+x^{2}+\left(4-x\right)^{2}=25$$

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

$$9+6x+x^{2}+16-8x+x^{2}=25$$

Add 9 and 16 to get 25.

$$25+6x+x^{2}-8x+x^{2}=25$$

Combine 6x and -8x to get -2x.

$$25-2x+x^{2}+x^{2}=25$$

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

$$25-2x+2x^{2}=25$$

Subtract 25 from both sides.

$$25-2x+2x^{2}-25=0$$

Subtract 25 from 25 to get 0.

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

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.

$$2x^{2}-2x=0$$

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

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

Take the square root of \left(-2\right)^{2}.

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

The opposite of -2 is 2.

$$x=\frac{2±2}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{2±2}{4}$$

Now solve the equation x=\frac{2±2}{4} when ± is plus. Add 2 to 2.

$$x=\frac{4}{4}$$

Divide 4 by 4.

$$x=1$$

Now solve the equation x=\frac{2±2}{4} when ± is minus. Subtract 2 from 2.

$$x=\frac{0}{4}$$

Divide 0 by 4.

$$x=0$$

The equation is now solved.

$$x=1 x=0$$

Steps for Completing the Square

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

$$9+6x+x^{2}+\left(4-x\right)^{2}=25$$

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

$$9+6x+x^{2}+16-8x+x^{2}=25$$

Add 9 and 16 to get 25.

$$25+6x+x^{2}-8x+x^{2}=25$$

Combine 6x and -8x to get -2x.

$$25-2x+x^{2}+x^{2}=25$$

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

$$25-2x+2x^{2}=25$$

Subtract 25 from both sides.

$$-2x+2x^{2}=25-25$$

Subtract 25 from 25 to get 0.

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

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$2x^{2}-2x=0$$

Divide both sides by 2.

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

Dividing by 2 undoes the multiplication by 2.

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

Divide -2 by 2.

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

Divide 0 by 2.

$$x^{2}-x=0$$

Divide -1, the coefficient of the x term, by 2 to get -\frac{1}{2}. Then add the square of -\frac{1}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -\frac{1}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-x+\frac{1}{4}=\frac{1}{4}$$

Factor x^{2}-x+\frac{1}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$x-\frac{1}{2}=\frac{1}{2} x-\frac{1}{2}=-\frac{1}{2}$$

Add \frac{1}{2} to both sides of the equation.

$$x=1 x=0$$