Solve x^2-13x+42=1

$$x ^ { 2 } - 13 x + 42 = 1$$

Solve for x

$$x = \frac{\sqrt{5} + 13}{2} \approx 7.618033989$$

$$x = \frac{13 - \sqrt{5}}{2} \approx 5.381966011$$

Steps Using the Quadratic Formula

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.

$$x^{2}-13x+42=1$$

Subtract 1 from both sides of the equation.

$$x^{2}-13x+42-1=1-1$$

Subtracting 1 from itself leaves 0.

$$x^{2}-13x+42-1=0$$

Subtract 1 from 42.

$$x^{2}-13x+41=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -13 for b, and 41 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-13\right)±\sqrt{\left(-13\right)^{2}-4\times 41}}{2}$$

Square -13.

$$x=\frac{-\left(-13\right)±\sqrt{169-4\times 41}}{2}$$

Multiply -4 times 41.

$$x=\frac{-\left(-13\right)±\sqrt{169-164}}{2}$$

Add 169 to -164.

$$x=\frac{-\left(-13\right)±\sqrt{5}}{2}$$

The opposite of -13 is 13.

$$x=\frac{13±\sqrt{5}}{2}$$

Now solve the equation x=\frac{13±\sqrt{5}}{2} when ± is plus. Add 13 to \sqrt{5}.

$$x=\frac{\sqrt{5}+13}{2}$$

Now solve the equation x=\frac{13±\sqrt{5}}{2} when ± is minus. Subtract \sqrt{5} from 13.

$$x=\frac{13-\sqrt{5}}{2}$$

The equation is now solved.

$$x=\frac{\sqrt{5}+13}{2} x=\frac{13-\sqrt{5}}{2}$$

Steps for Completing the Square

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.

$$x^{2}-13x+42=1$$

Subtract 42 from both sides of the equation.

$$x^{2}-13x+42-42=1-42$$

Subtracting 42 from itself leaves 0.

$$x^{2}-13x=1-42$$

Subtract 42 from 1.

$$x^{2}-13x=-41$$

Divide -13, the coefficient of the x term, by 2 to get -\frac{13}{2}. Then add the square of -\frac{13}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-13x+\left(-\frac{13}{2}\right)^{2}=-41+\left(-\frac{13}{2}\right)^{2}$$

Square -\frac{13}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-13x+\frac{169}{4}=-41+\frac{169}{4}$$

Add -41 to \frac{169}{4}.

$$x^{2}-13x+\frac{169}{4}=\frac{5}{4}$$

Factor x^{2}-13x+\frac{169}{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}.

$$\left(x-\frac{13}{2}\right)^{2}=\frac{5}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{13}{2}\right)^{2}}=\sqrt{\frac{5}{4}}$$

Simplify.

$$x-\frac{13}{2}=\frac{\sqrt{5}}{2} x-\frac{13}{2}=-\frac{\sqrt{5}}{2}$$

Add \frac{13}{2} to both sides of the equation.

$$x=\frac{\sqrt{5}+13}{2} x=\frac{13-\sqrt{5}}{2}$$