Solve x^2+x^2-11x-60=08

$$x ^ { 2 } + x ^ { 2 } - 11 x - 60 = 08$$

Solve for x

$$x = \frac{\sqrt{601} + 11}{4} \approx 8.878825336$$

$$x=\frac{11-\sqrt{601}}{4}\approx -3.378825336$$

Steps Using the Quadratic Formula

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

$$2x^{2}-11x-60=0\times 8$$

Multiply 0 and 8 to get 0.

$$2x^{2}-11x-60=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -11 for b, and -60 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-11\right)±\sqrt{\left(-11\right)^{2}-4\times 2\left(-60\right)}}{2\times 2}$$

Square -11.

$$x=\frac{-\left(-11\right)±\sqrt{121-4\times 2\left(-60\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-11\right)±\sqrt{121-8\left(-60\right)}}{2\times 2}$$

Multiply -8 times -60.

$$x=\frac{-\left(-11\right)±\sqrt{121+480}}{2\times 2}$$

Add 121 to 480.

$$x=\frac{-\left(-11\right)±\sqrt{601}}{2\times 2}$$

The opposite of -11 is 11.

$$x=\frac{11±\sqrt{601}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{11±\sqrt{601}}{4}$$

Now solve the equation x=\frac{11±\sqrt{601}}{4} when ± is plus. Add 11 to \sqrt{601}.

$$x=\frac{\sqrt{601}+11}{4}$$

Now solve the equation x=\frac{11±\sqrt{601}}{4} when ± is minus. Subtract \sqrt{601} from 11.

$$x=\frac{11-\sqrt{601}}{4}$$

The equation is now solved.

$$x=\frac{\sqrt{601}+11}{4} x=\frac{11-\sqrt{601}}{4}$$

Steps for Completing the Square

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

$$2x^{2}-11x-60=0\times 8$$

Multiply 0 and 8 to get 0.

$$2x^{2}-11x-60=0$$

Add 60 to both sides. Anything plus zero gives itself.

$$2x^{2}-11x=60$$

Divide both sides by 2.

$$\frac{2x^{2}-11x}{2}=\frac{60}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}-\frac{11}{2}x=\frac{60}{2}$$

Divide 60 by 2.

$$x^{2}-\frac{11}{2}x=30$$

Divide -\frac{11}{2}, the coefficient of the x term, by 2 to get -\frac{11}{4}. Then add the square of -\frac{11}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{11}{2}x+\left(-\frac{11}{4}\right)^{2}=30+\left(-\frac{11}{4}\right)^{2}$$

Square -\frac{11}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{11}{2}x+\frac{121}{16}=30+\frac{121}{16}$$

Add 30 to \frac{121}{16}.

$$x^{2}-\frac{11}{2}x+\frac{121}{16}=\frac{601}{16}$$

Factor x^{2}-\frac{11}{2}x+\frac{121}{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}.

$$\left(x-\frac{11}{4}\right)^{2}=\frac{601}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{11}{4}\right)^{2}}=\sqrt{\frac{601}{16}}$$

Simplify.

$$x-\frac{11}{4}=\frac{\sqrt{601}}{4} x-\frac{11}{4}=-\frac{\sqrt{601}}{4}$$

Add \frac{11}{4} to both sides of the equation.

$$x=\frac{\sqrt{601}+11}{4} x=\frac{11-\sqrt{601}}{4}$$