Solve 9x^2-49=0

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

Solve for x

$$x = \frac{7}{3} = 2\frac{1}{3} \approx 2.333333333$$

$$x = -\frac{7}{3} = -2\frac{1}{3} \approx -2.333333333$$

Steps Using Difference of Squares

Consider 9x^{2}-49. Rewrite 9x^{2}-49 as \left(3x\right)^{2}-7^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(3x-7\right)\left(3x+7\right)=0$$

To find equation solutions, solve 3x-7=0 and 3x+7=0.

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

Steps by Finding Square Root

Add 49 to both sides. Anything plus zero gives itself.

$$9x^{2}=49$$

Divide both sides by 9.

$$x^{2}=\frac{49}{9}$$

Take the square root of both sides of the equation.

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

Steps Using the Quadratic Formula

Quadratic equations like this one, with an x^{2} term but no x term, can still be solved using the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}, once they are put in standard form: ax^{2}+bx+c=0.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute 9 for a, 0 for b, and -49 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\times 9\left(-49\right)}}{2\times 9}$$

Square 0.

$$x=\frac{0±\sqrt{-4\times 9\left(-49\right)}}{2\times 9}$$

Multiply -4 times 9.

$$x=\frac{0±\sqrt{-36\left(-49\right)}}{2\times 9}$$

Multiply -36 times -49.

$$x=\frac{0±\sqrt{1764}}{2\times 9}$$

Take the square root of 1764.

$$x=\frac{0±42}{2\times 9}$$

Multiply 2 times 9.

$$x=\frac{0±42}{18}$$

Now solve the equation x=\frac{0±42}{18} when ± is plus. Reduce the fraction \frac{42}{18} to lowest terms by extracting and canceling out 6.

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

Now solve the equation x=\frac{0±42}{18} when ± is minus. Reduce the fraction \frac{-42}{18} to lowest terms by extracting and canceling out 6.

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

The equation is now solved.

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