Solve 16x^2-25{left(4/3x-16/3right)}^2=400

$$16 { x }^{ 2 } -25 { \left( \frac{ 4 }{ 3 } x- \frac{ 16 }{ 3 } \right) }^{ 2 } =400$$

Solve for x

$$x = \frac{25}{4} = 6\frac{1}{4} = 6.25$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\frac{4}{3}x-\frac{16}{3}\right)^{2}.

$$16x^{2}-25\left(\frac{16}{9}x^{2}-\frac{128}{9}x+\frac{256}{9}\right)=400$$

Use the distributive property to multiply -25 by \frac{16}{9}x^{2}-\frac{128}{9}x+\frac{256}{9}.

$$16x^{2}-\frac{400}{9}x^{2}+\frac{3200}{9}x-\frac{6400}{9}=400$$

Combine 16x^{2} and -\frac{400}{9}x^{2} to get -\frac{256}{9}x^{2}.

$$-\frac{256}{9}x^{2}+\frac{3200}{9}x-\frac{6400}{9}=400$$

Subtract 400 from both sides.

$$-\frac{256}{9}x^{2}+\frac{3200}{9}x-\frac{6400}{9}-400=0$$

Subtract 400 from -\frac{6400}{9} to get -\frac{10000}{9}.

$$-\frac{256}{9}x^{2}+\frac{3200}{9}x-\frac{10000}{9}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -\frac{256}{9} for a, \frac{3200}{9} for b, and -\frac{10000}{9} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\frac{3200}{9}±\sqrt{\left(\frac{3200}{9}\right)^{2}-4\left(-\frac{256}{9}\right)\left(-\frac{10000}{9}\right)}}{2\left(-\frac{256}{9}\right)}$$

Square \frac{3200}{9} by squaring both the numerator and the denominator of the fraction.

$$x=\frac{-\frac{3200}{9}±\sqrt{\frac{10240000}{81}-4\left(-\frac{256}{9}\right)\left(-\frac{10000}{9}\right)}}{2\left(-\frac{256}{9}\right)}$$

Multiply -4 times -\frac{256}{9}.

$$x=\frac{-\frac{3200}{9}±\sqrt{\frac{10240000}{81}+\frac{1024}{9}\left(-\frac{10000}{9}\right)}}{2\left(-\frac{256}{9}\right)}$$

Multiply \frac{1024}{9} times -\frac{10000}{9} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$x=\frac{-\frac{3200}{9}±\sqrt{\frac{10240000-10240000}{81}}}{2\left(-\frac{256}{9}\right)}$$

Add \frac{10240000}{81} to -\frac{10240000}{81} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{-\frac{3200}{9}±\sqrt{0}}{2\left(-\frac{256}{9}\right)}$$

Take the square root of 0.

$$x=-\frac{\frac{3200}{9}}{2\left(-\frac{256}{9}\right)}$$

Multiply 2 times -\frac{256}{9}.

$$x=-\frac{\frac{3200}{9}}{-\frac{512}{9}}$$

Divide -\frac{3200}{9} by -\frac{512}{9} by multiplying -\frac{3200}{9} by the reciprocal of -\frac{512}{9}.

$$x=\frac{25}{4}$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\frac{4}{3}x-\frac{16}{3}\right)^{2}.

$$16x^{2}-25\left(\frac{16}{9}x^{2}-\frac{128}{9}x+\frac{256}{9}\right)=400$$

Use the distributive property to multiply -25 by \frac{16}{9}x^{2}-\frac{128}{9}x+\frac{256}{9}.

$$16x^{2}-\frac{400}{9}x^{2}+\frac{3200}{9}x-\frac{6400}{9}=400$$

Combine 16x^{2} and -\frac{400}{9}x^{2} to get -\frac{256}{9}x^{2}.

$$-\frac{256}{9}x^{2}+\frac{3200}{9}x-\frac{6400}{9}=400$$

Add \frac{6400}{9} to both sides.

$$-\frac{256}{9}x^{2}+\frac{3200}{9}x=400+\frac{6400}{9}$$

Add 400 and \frac{6400}{9} to get \frac{10000}{9}.

$$-\frac{256}{9}x^{2}+\frac{3200}{9}x=\frac{10000}{9}$$

Divide both sides of the equation by -\frac{256}{9}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{-\frac{256}{9}x^{2}+\frac{3200}{9}x}{-\frac{256}{9}}=\frac{\frac{10000}{9}}{-\frac{256}{9}}$$

Dividing by -\frac{256}{9} undoes the multiplication by -\frac{256}{9}.

$$x^{2}+\frac{\frac{3200}{9}}{-\frac{256}{9}}x=\frac{\frac{10000}{9}}{-\frac{256}{9}}$$

Divide \frac{3200}{9} by -\frac{256}{9} by multiplying \frac{3200}{9} by the reciprocal of -\frac{256}{9}.

$$x^{2}-\frac{25}{2}x=\frac{\frac{10000}{9}}{-\frac{256}{9}}$$

Divide \frac{10000}{9} by -\frac{256}{9} by multiplying \frac{10000}{9} by the reciprocal of -\frac{256}{9}.

$$x^{2}-\frac{25}{2}x=-\frac{625}{16}$$

Divide -\frac{25}{2}, the coefficient of the x term, by 2 to get -\frac{25}{4}. Then add the square of -\frac{25}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{25}{2}x+\left(-\frac{25}{4}\right)^{2}=-\frac{625}{16}+\left(-\frac{25}{4}\right)^{2}$$

Square -\frac{25}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{25}{2}x+\frac{625}{16}=\frac{-625+625}{16}$$

Add -\frac{625}{16} to \frac{625}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{25}{2}x+\frac{625}{16}=0$$

Factor x^{2}-\frac{25}{2}x+\frac{625}{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}.

$$\left(x-\frac{25}{4}\right)^{2}=0$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{25}{4}\right)^{2}}=\sqrt{0}$$

Simplify.

$$x-\frac{25}{4}=0 x-\frac{25}{4}=0$$

Add \frac{25}{4} to both sides of the equation.

$$x=\frac{25}{4} x=\frac{25}{4}$$

The equation is now solved. Solutions are the same.

$$x=\frac{25}{4}$$