Solve 4+x^2+1+left(x-3right)^2=25

$$4+ { x }^{ 2 } +1+ { \left(x-3 \right) }^{ 2 } =25$$

Solve for x

$$x = \frac{\sqrt{31} + 3}{2} \approx 4.283882181$$

$$x=\frac{3-\sqrt{31}}{2}\approx -1.283882181$$

Steps Using the Quadratic Formula

Add 4 and 1 to get 5.

$$5+x^{2}+\left(x-3\right)^{2}=25$$

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

$$5+x^{2}+x^{2}-6x+9=25$$

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

$$5+2x^{2}-6x+9=25$$

Add 5 and 9 to get 14.

$$14+2x^{2}-6x=25$$

Subtract 25 from both sides.

$$14+2x^{2}-6x-25=0$$

Subtract 25 from 14 to get -11.

$$-11+2x^{2}-6x=0$$

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.

$$2x^{2}-6x-11=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -6 for b, and -11 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\times 2\left(-11\right)}}{2\times 2}$$

Square -6.

$$x=\frac{-\left(-6\right)±\sqrt{36-4\times 2\left(-11\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-6\right)±\sqrt{36-8\left(-11\right)}}{2\times 2}$$

Multiply -8 times -11.

$$x=\frac{-\left(-6\right)±\sqrt{36+88}}{2\times 2}$$

Add 36 to 88.

$$x=\frac{-\left(-6\right)±\sqrt{124}}{2\times 2}$$

Take the square root of 124.

$$x=\frac{-\left(-6\right)±2\sqrt{31}}{2\times 2}$$

The opposite of -6 is 6.

$$x=\frac{6±2\sqrt{31}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{6±2\sqrt{31}}{4}$$

Now solve the equation x=\frac{6±2\sqrt{31}}{4} when ± is plus. Add 6 to 2\sqrt{31}.

$$x=\frac{2\sqrt{31}+6}{4}$$

Divide 6+2\sqrt{31} by 4.

$$x=\frac{\sqrt{31}+3}{2}$$

Now solve the equation x=\frac{6±2\sqrt{31}}{4} when ± is minus. Subtract 2\sqrt{31} from 6.

$$x=\frac{6-2\sqrt{31}}{4}$$

Divide 6-2\sqrt{31} by 4.

$$x=\frac{3-\sqrt{31}}{2}$$

The equation is now solved.

$$x=\frac{\sqrt{31}+3}{2} x=\frac{3-\sqrt{31}}{2}$$

Steps for Completing the Square

Add 4 and 1 to get 5.

$$5+x^{2}+\left(x-3\right)^{2}=25$$

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

$$5+x^{2}+x^{2}-6x+9=25$$

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

$$5+2x^{2}-6x+9=25$$

Add 5 and 9 to get 14.

$$14+2x^{2}-6x=25$$

Subtract 14 from both sides.

$$2x^{2}-6x=25-14$$

Subtract 14 from 25 to get 11.

$$2x^{2}-6x=11$$

Divide both sides by 2.

$$\frac{2x^{2}-6x}{2}=\frac{11}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\left(-\frac{6}{2}\right)x=\frac{11}{2}$$

Divide -6 by 2.

$$x^{2}-3x=\frac{11}{2}$$

Divide -3, the coefficient of the x term, by 2 to get -\frac{3}{2}. Then add the square of -\frac{3}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-3x+\left(-\frac{3}{2}\right)^{2}=\frac{11}{2}+\left(-\frac{3}{2}\right)^{2}$$

Square -\frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-3x+\frac{9}{4}=\frac{11}{2}+\frac{9}{4}$$

Add \frac{11}{2} to \frac{9}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-3x+\frac{9}{4}=\frac{31}{4}$$

Factor x^{2}-3x+\frac{9}{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}.

$$\left(x-\frac{3}{2}\right)^{2}=\frac{31}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{2}\right)^{2}}=\sqrt{\frac{31}{4}}$$

Simplify.

$$x-\frac{3}{2}=\frac{\sqrt{31}}{2} x-\frac{3}{2}=-\frac{\sqrt{31}}{2}$$

Add \frac{3}{2} to both sides of the equation.

$$x=\frac{\sqrt{31}+3}{2} x=\frac{3-\sqrt{31}}{2}$$