Solve x^2=4x^2-32x+48

$${ x }^{ 2 } = 4 { x }^{ 2 } -32x+48$$

Solve for x

$$x = \frac{4 \sqrt{7} + 16}{3} \approx 8.861001748$$

$$x = \frac{16 - 4 \sqrt{7}}{3} \approx 1.805664919$$

Steps Using the Quadratic Formula

Subtract 4x^{2} from both sides.

$$x^{2}-4x^{2}=-32x+48$$

Combine x^{2} and -4x^{2} to get -3x^{2}.

$$-3x^{2}=-32x+48$$

Add 32x to both sides.

$$-3x^{2}+32x=48$$

Subtract 48 from both sides.

$$-3x^{2}+32x-48=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -3 for a, 32 for b, and -48 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-32±\sqrt{32^{2}-4\left(-3\right)\left(-48\right)}}{2\left(-3\right)}$$

Square 32.

$$x=\frac{-32±\sqrt{1024-4\left(-3\right)\left(-48\right)}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$x=\frac{-32±\sqrt{1024+12\left(-48\right)}}{2\left(-3\right)}$$

Multiply 12 times -48.

$$x=\frac{-32±\sqrt{1024-576}}{2\left(-3\right)}$$

Add 1024 to -576.

$$x=\frac{-32±\sqrt{448}}{2\left(-3\right)}$$

Take the square root of 448.

$$x=\frac{-32±8\sqrt{7}}{2\left(-3\right)}$$

Multiply 2 times -3.

$$x=\frac{-32±8\sqrt{7}}{-6}$$

Now solve the equation x=\frac{-32±8\sqrt{7}}{-6} when ± is plus. Add -32 to 8\sqrt{7}.

$$x=\frac{8\sqrt{7}-32}{-6}$$

Divide -32+8\sqrt{7} by -6.

$$x=\frac{16-4\sqrt{7}}{3}$$

Now solve the equation x=\frac{-32±8\sqrt{7}}{-6} when ± is minus. Subtract 8\sqrt{7} from -32.

$$x=\frac{-8\sqrt{7}-32}{-6}$$

Divide -32-8\sqrt{7} by -6.

$$x=\frac{4\sqrt{7}+16}{3}$$

The equation is now solved.

$$x=\frac{16-4\sqrt{7}}{3} x=\frac{4\sqrt{7}+16}{3}$$

Steps for Completing the Square

Subtract 4x^{2} from both sides.

$$x^{2}-4x^{2}=-32x+48$$

Combine x^{2} and -4x^{2} to get -3x^{2}.

$$-3x^{2}=-32x+48$$

Add 32x to both sides.

$$-3x^{2}+32x=48$$

Divide both sides by -3.

$$\frac{-3x^{2}+32x}{-3}=\frac{48}{-3}$$

Dividing by -3 undoes the multiplication by -3.

$$x^{2}+\frac{32}{-3}x=\frac{48}{-3}$$

Divide 32 by -3.

$$x^{2}-\frac{32}{3}x=\frac{48}{-3}$$

Divide 48 by -3.

$$x^{2}-\frac{32}{3}x=-16$$

Divide -\frac{32}{3}, the coefficient of the x term, by 2 to get -\frac{16}{3}. Then add the square of -\frac{16}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{32}{3}x+\left(-\frac{16}{3}\right)^{2}=-16+\left(-\frac{16}{3}\right)^{2}$$

Square -\frac{16}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{32}{3}x+\frac{256}{9}=-16+\frac{256}{9}$$

Add -16 to \frac{256}{9}.

$$x^{2}-\frac{32}{3}x+\frac{256}{9}=\frac{112}{9}$$

Factor x^{2}-\frac{32}{3}x+\frac{256}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{16}{3}\right)^{2}=\frac{112}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{16}{3}\right)^{2}}=\sqrt{\frac{112}{9}}$$

Simplify.

$$x-\frac{16}{3}=\frac{4\sqrt{7}}{3} x-\frac{16}{3}=-\frac{4\sqrt{7}}{3}$$

Add \frac{16}{3} to both sides of the equation.

$$x=\frac{4\sqrt{7}+16}{3} x=\frac{16-4\sqrt{7}}{3}$$