Solve for a 9x^2-24x+16=(ax-b)^2

$9 x^{2} - 24 x + 16 = {(a x - b)}^{2}$

Step 1

Rewrite the equation as ${(a x - b)}^{2} = 9 x^{2} - 24 x + 16$ .

${(a x - b)}^{2} = 9 x^{2} - 24 x + 16$

Step 2

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$a x - b = \pm \sqrt{9 x^{2} - 24 x + 16}$

Step 3

Simplify $\pm \sqrt{9 x^{2} - 24 x + 16}$ .

Factor using the perfect square rule.

Rewrite $9 x^{2}$ as ${(3 x)}^{2}$ .

$a x - b = \pm \sqrt{{(3 x)}^{2} - 24 x + 16}$

Rewrite $16$ as $4^{2}$ .

$a x - b = \pm \sqrt{{(3 x)}^{2} - 24 x + 4^{2}}$

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$24 x = 2 \cdot (3 x) \cdot 4$

Rewrite the polynomial.

$a x - b = \pm \sqrt{{(3 x)}^{2} - 2 \cdot (3 x) \cdot 4 + 4^{2}}$

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = 3 x$ and $b = 4$ .

$a x - b = \pm \sqrt{{(3 x - 4)}^{2}}$

Pull terms out from under the radical, assuming positive real numbers.

$a x - b = \pm (3 x - 4)$

Step 4

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

$a x - b = 3 x - 4$

Add $b$ to both sides of the equation.

$a x = 3 x - 4 + b$

Divide each term in $a x = 3 x - 4 + b$ by $x$ and simplify.

Divide each term in $a x = 3 x - 4 + b$ by $x$ .

$\frac{a x}{x} = \frac{3 x}{x} + \frac{- 4}{x} + \frac{b}{x}$

Simplify the left side.

Cancel the common factor of $x$ .

Cancel the common factor.

$\frac{a x}{x} = \frac{3 x}{x} + \frac{- 4}{x} + \frac{b}{x}$

Divide $a$ by $1$ .

$a = \frac{3 x}{x} + \frac{- 4}{x} + \frac{b}{x}$

Simplify the right side.

Simplify each term.

Cancel the common factor of $x$ .

Cancel the common factor.

$a = \frac{3 x}{x} + \frac{- 4}{x} + \frac{b}{x}$

Divide $3$ by $1$ .

$a = 3 + \frac{- 4}{x} + \frac{b}{x}$

Move the negative in front of the fraction.

$a = 3 - \frac{4}{x} + \frac{b}{x}$

Next, use the negative value of the $\pm$ to find the second solution.

$a x - b = - (3 x - 4)$

Simplify $- (3 x - 4)$ .

Rewrite.

$a x - b = 0 + 0 - (3 x - 4)$

Simplify by adding zeros.

$a x - b = - (3 x - 4)$

Apply the distributive property.

$a x - b = - (3 x) - - 4$

Multiply.

Multiply $3$ by $- 1$ .

$a x - b = - 3 x - - 4$

Multiply $- 1$ by $- 4$ .

$a x - b = - 3 x + 4$

Add $b$ to both sides of the equation.

$a x = - 3 x + 4 + b$

Divide each term in $a x = - 3 x + 4 + b$ by $x$ and simplify.

Divide each term in $a x = - 3 x + 4 + b$ by $x$ .

$\frac{a x}{x} = \frac{- 3 x}{x} + \frac{4}{x} + \frac{b}{x}$

Simplify the left side.

Cancel the common factor of $x$ .

Cancel the common factor.

$\frac{a x}{x} = \frac{- 3 x}{x} + \frac{4}{x} + \frac{b}{x}$

Divide $a$ by $1$ .

$a = \frac{- 3 x}{x} + \frac{4}{x} + \frac{b}{x}$

Simplify the right side.

Cancel the common factor of $x$ .

Cancel the common factor.

$a = \frac{- 3 x}{x} + \frac{4}{x} + \frac{b}{x}$

Divide $- 3$ by $1$ .

$a = - 3 + \frac{4}{x} + \frac{b}{x}$

The complete solution is the result of both the positive and negative portions of the solution.

$a = 3 - \frac{4}{x} + \frac{b}{x}$

$a = - 3 + \frac{4}{x} + \frac{b}{x}$

$a = 3 - \frac{4}{x} + \frac{b}{x}$

$a = - 3 + \frac{4}{x} + \frac{b}{x}$

Solve for a 9x^2-24x+16=(ax-b)^2

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