Solve 7x=4x^2-1

$$7 x = 4 x ^ { 2 } - 1$$

Solve for x

$$x = \frac{\sqrt{65} + 7}{8} \approx 1.882782219$$

$$x=\frac{7-\sqrt{65}}{8}\approx -0.132782219$$

Steps Using the Quadratic Formula

Subtract 4x^{2} from both sides.

$$7x-4x^{2}=-1$$

Add 1 to both sides.

$$7x-4x^{2}+1=0$$

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.

$$-4x^{2}+7x+1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -4 for a, 7 for b, and 1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-7±\sqrt{7^{2}-4\left(-4\right)}}{2\left(-4\right)}$$

Square 7.

$$x=\frac{-7±\sqrt{49-4\left(-4\right)}}{2\left(-4\right)}$$

Multiply -4 times -4.

$$x=\frac{-7±\sqrt{49+16}}{2\left(-4\right)}$$

Add 49 to 16.

$$x=\frac{-7±\sqrt{65}}{2\left(-4\right)}$$

Multiply 2 times -4.

$$x=\frac{-7±\sqrt{65}}{-8}$$

Now solve the equation x=\frac{-7±\sqrt{65}}{-8} when ± is plus. Add -7 to \sqrt{65}.

$$x=\frac{\sqrt{65}-7}{-8}$$

Divide -7+\sqrt{65} by -8.

$$x=\frac{7-\sqrt{65}}{8}$$

Now solve the equation x=\frac{-7±\sqrt{65}}{-8} when ± is minus. Subtract \sqrt{65} from -7.

$$x=\frac{-\sqrt{65}-7}{-8}$$

Divide -7-\sqrt{65} by -8.

$$x=\frac{\sqrt{65}+7}{8}$$

The equation is now solved.

$$x=\frac{7-\sqrt{65}}{8} x=\frac{\sqrt{65}+7}{8}$$

Steps for Completing the Square

Subtract 4x^{2} from both sides.

$$7x-4x^{2}=-1$$

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.

$$-4x^{2}+7x=-1$$

Divide both sides by -4.

$$\frac{-4x^{2}+7x}{-4}=-\frac{1}{-4}$$

Dividing by -4 undoes the multiplication by -4.

$$x^{2}+\frac{7}{-4}x=-\frac{1}{-4}$$

Divide 7 by -4.

$$x^{2}-\frac{7}{4}x=-\frac{1}{-4}$$

Divide -1 by -4.

$$x^{2}-\frac{7}{4}x=\frac{1}{4}$$

Divide -\frac{7}{4}, the coefficient of the x term, by 2 to get -\frac{7}{8}. Then add the square of -\frac{7}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{7}{4}x+\left(-\frac{7}{8}\right)^{2}=\frac{1}{4}+\left(-\frac{7}{8}\right)^{2}$$

Square -\frac{7}{8} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{7}{4}x+\frac{49}{64}=\frac{1}{4}+\frac{49}{64}$$

Add \frac{1}{4} to \frac{49}{64} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{7}{4}x+\frac{49}{64}=\frac{65}{64}$$

Factor x^{2}-\frac{7}{4}x+\frac{49}{64}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{7}{8}\right)^{2}=\frac{65}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{7}{8}\right)^{2}}=\sqrt{\frac{65}{64}}$$

Simplify.

$$x-\frac{7}{8}=\frac{\sqrt{65}}{8} x-\frac{7}{8}=-\frac{\sqrt{65}}{8}$$

Add \frac{7}{8} to both sides of the equation.

$$x=\frac{\sqrt{65}+7}{8} x=\frac{7-\sqrt{65}}{8}$$