Solve Using the Quadratic Formula 4x^2+nx=-9

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

Step 1

Add $9$ to both sides of the equation.

$4 x^{2} + n x + 9 = 0$

Step 2

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 3

Substitute the values $a = 4$ , $b = n$ , and $c = 9$ into the quadratic formula and solve for $x$ .

$\frac{- n \pm \sqrt{n^{2} - 4 \cdot (4 \cdot 9)}}{2 \cdot 4}$

Step 4

Simplify.

Simplify the numerator.

Rewrite $4 \cdot 4 \cdot 9$ as ${(2 \cdot 2 \cdot 3)}^{2}$ .

$x = \frac{- n \pm \sqrt{n^{2} - {(2 \cdot 2 \cdot 3)}^{2}}}{2 \cdot 4}$

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = n$ and $b = 2 \cdot 2 \cdot 3$ .

$x = \frac{- n \pm \sqrt{(n + 2 \cdot 2 \cdot 3) (n - (2 \cdot 2 \cdot 3))}}{2 \cdot 4}$

Simplify.

Multiply $2 \cdot 2 \cdot 3$ .

Multiply $2$ by $2$ .

$x = \frac{- n \pm \sqrt{(n + 4 \cdot 3) (n - (2 \cdot 2 \cdot 3))}}{2 \cdot 4}$

Multiply $4$ by $3$ .

$x = \frac{- n \pm \sqrt{(n + 12) (n - (2 \cdot 2 \cdot 3))}}{2 \cdot 4}$

Multiply $- (2 \cdot 2 \cdot 3)$ .

Multiply $2$ by $2$ .

$x = \frac{- n \pm \sqrt{(n + 12) (n - (4 \cdot 3))}}{2 \cdot 4}$

Multiply $4$ by $3$ .

$x = \frac{- n \pm \sqrt{(n + 12) (n - 1 \cdot 12)}}{2 \cdot 4}$

Multiply $- 1$ by $12$ .

$x = \frac{- n \pm \sqrt{(n + 12) (n - 12)}}{2 \cdot 4}$

Multiply $2$ by $4$ .

$x = \frac{- n \pm \sqrt{(n + 12) (n - 12)}}{8}$

Solve Using the Quadratic Formula 4x^2+nx=-9

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