Solve 16(x^2-9)=25x^2-90x+81

$$16 ( x ^ { 2 } - 9 ) = 25 x ^ { 2 } - 90 x + 81$$

Solve for x

$$x=5$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 16 by x^{2}-9.

$$16x^{2}-144=25x^{2}-90x+81$$

Subtract 25x^{2} from both sides.

$$16x^{2}-144-25x^{2}=-90x+81$$

Combine 16x^{2} and -25x^{2} to get -9x^{2}.

$$-9x^{2}-144=-90x+81$$

Add 90x to both sides.

$$-9x^{2}-144+90x=81$$

Subtract 81 from both sides.

$$-9x^{2}-144+90x-81=0$$

Subtract 81 from -144 to get -225.

$$-9x^{2}-225+90x=0$$

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.

$$-9x^{2}+90x-225=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -9 for a, 90 for b, and -225 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-90±\sqrt{90^{2}-4\left(-9\right)\left(-225\right)}}{2\left(-9\right)}$$

Square 90.

$$x=\frac{-90±\sqrt{8100-4\left(-9\right)\left(-225\right)}}{2\left(-9\right)}$$

Multiply -4 times -9.

$$x=\frac{-90±\sqrt{8100+36\left(-225\right)}}{2\left(-9\right)}$$

Multiply 36 times -225.

$$x=\frac{-90±\sqrt{8100-8100}}{2\left(-9\right)}$$

Add 8100 to -8100.

$$x=\frac{-90±\sqrt{0}}{2\left(-9\right)}$$

Take the square root of 0.

$$x=-\frac{90}{2\left(-9\right)}$$

Multiply 2 times -9.

$$x=-\frac{90}{-18}$$

Divide -90 by -18.

$$x=5$$

Steps for Completing the Square

Use the distributive property to multiply 16 by x^{2}-9.

$$16x^{2}-144=25x^{2}-90x+81$$

Subtract 25x^{2} from both sides.

$$16x^{2}-144-25x^{2}=-90x+81$$

Combine 16x^{2} and -25x^{2} to get -9x^{2}.

$$-9x^{2}-144=-90x+81$$

Add 90x to both sides.

$$-9x^{2}-144+90x=81$$

Add 144 to both sides.

$$-9x^{2}+90x=81+144$$

Add 81 and 144 to get 225.

$$-9x^{2}+90x=225$$

Divide both sides by -9.

$$\frac{-9x^{2}+90x}{-9}=\frac{225}{-9}$$

Dividing by -9 undoes the multiplication by -9.

$$x^{2}+\frac{90}{-9}x=\frac{225}{-9}$$

Divide 90 by -9.

$$x^{2}-10x=\frac{225}{-9}$$

Divide 225 by -9.

$$x^{2}-10x=-25$$

Divide -10, the coefficient of the x term, by 2 to get -5. Then add the square of -5 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-10x+\left(-5\right)^{2}=-25+\left(-5\right)^{2}$$

Square -5.

$$x^{2}-10x+25=-25+25$$

Add -25 to 25.

$$x^{2}-10x+25=0$$

Factor x^{2}-10x+25. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-5\right)^{2}=0$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-5\right)^{2}}=\sqrt{0}$$

Simplify.

$$x-5=0 x-5=0$$

Add 5 to both sides of the equation.

$$x=5 x=5$$

The equation is now solved. Solutions are the same.

$$x=5$$