Solve Using the Quadratic Formula 49x^2-42x+9=0

$49 x^{2} - 42 x + 9 = 0$

Step 1

Use the quadratic formula to find the solutions.

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

Step 2

Substitute the values $a = 49$ , $b = - 42$ , and $c = 9$ into the quadratic formula and solve for $x$ .

$\frac{42 \pm \sqrt{{(- 42)}^{2} - 4 \cdot (49 \cdot 9)}}{2 \cdot 49}$

Step 3

Simplify.

Simplify the numerator.

Raise $- 42$ to the power of $2$ .

$x = \frac{42 \pm \sqrt{1764 - 4 \cdot 49 \cdot 9}}{2 \cdot 49}$

Multiply $- 4 \cdot 49 \cdot 9$ .

Multiply $- 4$ by $49$ .

$x = \frac{42 \pm \sqrt{1764 - 196 \cdot 9}}{2 \cdot 49}$

Multiply $- 196$ by $9$ .

$x = \frac{42 \pm \sqrt{1764 - 1764}}{2 \cdot 49}$

Subtract $1764$ from $1764$ .

$x = \frac{42 \pm \sqrt{0}}{2 \cdot 49}$

Rewrite $0$ as $0^{2}$ .

$x = \frac{42 \pm \sqrt{0^{2}}}{2 \cdot 49}$

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

$x = \frac{42 \pm 0}{2 \cdot 49}$

$42$ plus or minus $0$ is $42$ .

$x = \frac{42}{2 \cdot 49}$

Multiply $2$ by $49$ .

$x = \frac{42}{98}$

Cancel the common factor of $42$ and $98$ .

Factor $14$ out of $42$ .

$x = \frac{14 (3)}{98}$

Cancel the common factors.

Factor $14$ out of $98$ .

$x = \frac{14 \cdot 3}{14 \cdot 7}$

Cancel the common factor.

$x = \frac{14 \cdot 3}{14 \cdot 7}$

Rewrite the expression.

$x = \frac{3}{7}$

Step 4

The final answer is the combination of both solutions.

$x = \frac{3}{7}$ Double roots