Solve 5x^2-1=3x^2+9x

$$5 x ^ { 2 } - 1 = 3 x ^ { 2 } + 9 x$$

Solve for x

$$x = \frac{\sqrt{89} + 9}{4} \approx 4.608495283$$

$$x=\frac{9-\sqrt{89}}{4}\approx -0.108495283$$

Steps Using the Quadratic Formula

Subtract 3x^{2} from both sides.

$$5x^{2}-1-3x^{2}=9x$$

Combine 5x^{2} and -3x^{2} to get 2x^{2}.

$$2x^{2}-1=9x$$

Subtract 9x from both sides.

$$2x^{2}-1-9x=0$$

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.

$$2x^{2}-9x-1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -9 for b, and -1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-9\right)±\sqrt{\left(-9\right)^{2}-4\times 2\left(-1\right)}}{2\times 2}$$

Square -9.

$$x=\frac{-\left(-9\right)±\sqrt{81-4\times 2\left(-1\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-9\right)±\sqrt{81-8\left(-1\right)}}{2\times 2}$$

Multiply -8 times -1.

$$x=\frac{-\left(-9\right)±\sqrt{81+8}}{2\times 2}$$

Add 81 to 8.

$$x=\frac{-\left(-9\right)±\sqrt{89}}{2\times 2}$$

The opposite of -9 is 9.

$$x=\frac{9±\sqrt{89}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{9±\sqrt{89}}{4}$$

Now solve the equation x=\frac{9±\sqrt{89}}{4} when ± is plus. Add 9 to \sqrt{89}.

$$x=\frac{\sqrt{89}+9}{4}$$

Now solve the equation x=\frac{9±\sqrt{89}}{4} when ± is minus. Subtract \sqrt{89} from 9.

$$x=\frac{9-\sqrt{89}}{4}$$

The equation is now solved.

$$x=\frac{\sqrt{89}+9}{4} x=\frac{9-\sqrt{89}}{4}$$

Steps for Completing the Square

Subtract 3x^{2} from both sides.

$$5x^{2}-1-3x^{2}=9x$$

Combine 5x^{2} and -3x^{2} to get 2x^{2}.

$$2x^{2}-1=9x$$

Subtract 9x from both sides.

$$2x^{2}-1-9x=0$$

Add 1 to both sides. Anything plus zero gives itself.

$$2x^{2}-9x=1$$

Divide both sides by 2.

$$\frac{2x^{2}-9x}{2}=\frac{1}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}-\frac{9}{2}x=\frac{1}{2}$$

Divide -\frac{9}{2}, the coefficient of the x term, by 2 to get -\frac{9}{4}. Then add the square of -\frac{9}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{9}{2}x+\left(-\frac{9}{4}\right)^{2}=\frac{1}{2}+\left(-\frac{9}{4}\right)^{2}$$

Square -\frac{9}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{9}{2}x+\frac{81}{16}=\frac{1}{2}+\frac{81}{16}$$

Add \frac{1}{2} to \frac{81}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{9}{2}x+\frac{81}{16}=\frac{89}{16}$$

Factor x^{2}-\frac{9}{2}x+\frac{81}{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}.

$$\left(x-\frac{9}{4}\right)^{2}=\frac{89}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{9}{4}\right)^{2}}=\sqrt{\frac{89}{16}}$$

Simplify.

$$x-\frac{9}{4}=\frac{\sqrt{89}}{4} x-\frac{9}{4}=-\frac{\sqrt{89}}{4}$$

Add \frac{9}{4} to both sides of the equation.

$$x=\frac{\sqrt{89}+9}{4} x=\frac{9-\sqrt{89}}{4}$$