Solve (9-x)*4x=16

$$(9-x) \times 4x=16$$

Solve for x

$$x = \frac{\sqrt{65} + 9}{2} \approx 8.531128874$$

$$x=\frac{9-\sqrt{65}}{2}\approx 0.468871126$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 9-x by 4.

$$\left(36-4x\right)x=16$$

Use the distributive property to multiply 36-4x by x.

$$36x-4x^{2}=16$$

Subtract 16 from both sides.

$$36x-4x^{2}-16=0$$

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.

$$-4x^{2}+36x-16=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -4 for a, 36 for b, and -16 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-36±\sqrt{36^{2}-4\left(-4\right)\left(-16\right)}}{2\left(-4\right)}$$

Square 36.

$$x=\frac{-36±\sqrt{1296-4\left(-4\right)\left(-16\right)}}{2\left(-4\right)}$$

Multiply -4 times -4.

$$x=\frac{-36±\sqrt{1296+16\left(-16\right)}}{2\left(-4\right)}$$

Multiply 16 times -16.

$$x=\frac{-36±\sqrt{1296-256}}{2\left(-4\right)}$$

Add 1296 to -256.

$$x=\frac{-36±\sqrt{1040}}{2\left(-4\right)}$$

Take the square root of 1040.

$$x=\frac{-36±4\sqrt{65}}{2\left(-4\right)}$$

Multiply 2 times -4.

$$x=\frac{-36±4\sqrt{65}}{-8}$$

Now solve the equation x=\frac{-36±4\sqrt{65}}{-8} when ± is plus. Add -36 to 4\sqrt{65}.

$$x=\frac{4\sqrt{65}-36}{-8}$$

Divide -36+4\sqrt{65} by -8.

$$x=\frac{9-\sqrt{65}}{2}$$

Now solve the equation x=\frac{-36±4\sqrt{65}}{-8} when ± is minus. Subtract 4\sqrt{65} from -36.

$$x=\frac{-4\sqrt{65}-36}{-8}$$

Divide -36-4\sqrt{65} by -8.

$$x=\frac{\sqrt{65}+9}{2}$$

The equation is now solved.

$$x=\frac{9-\sqrt{65}}{2} x=\frac{\sqrt{65}+9}{2}$$

Steps for Completing the Square

Use the distributive property to multiply 9-x by 4.

$$\left(36-4x\right)x=16$$

Use the distributive property to multiply 36-4x by x.

$$36x-4x^{2}=16$$

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.

$$-4x^{2}+36x=16$$

Divide both sides by -4.

$$\frac{-4x^{2}+36x}{-4}=\frac{16}{-4}$$

Dividing by -4 undoes the multiplication by -4.

$$x^{2}+\frac{36}{-4}x=\frac{16}{-4}$$

Divide 36 by -4.

$$x^{2}-9x=\frac{16}{-4}$$

Divide 16 by -4.

$$x^{2}-9x=-4$$

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

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

Square -\frac{9}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-9x+\frac{81}{4}=-4+\frac{81}{4}$$

Add -4 to \frac{81}{4}.

$$x^{2}-9x+\frac{81}{4}=\frac{65}{4}$$

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

$$\left(x-\frac{9}{2}\right)^{2}=\frac{65}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{9}{2}\right)^{2}}=\sqrt{\frac{65}{4}}$$

Simplify.

$$x-\frac{9}{2}=\frac{\sqrt{65}}{2} x-\frac{9}{2}=-\frac{\sqrt{65}}{2}$$

Add \frac{9}{2} to both sides of the equation.

$$x=\frac{\sqrt{65}+9}{2} x=\frac{9-\sqrt{65}}{2}$$