Solve 6+x^2=sqrt{49/81}

$$6 + x ^ { 2 } = \sqrt { \frac { 49 } { 81 } }$$

Solve for x (complex solution)

$$x=-\frac{\sqrt{47}i}{3}\approx -0-2.2852182i$$

$$x=\frac{\sqrt{47}i}{3}\approx 2.2852182i$$

Steps by Finding Square Root

Rewrite the square root of the division \frac{49}{81} as the division of square roots \frac{\sqrt{49}}{\sqrt{81}}. Take the square root of both numerator and denominator.

$$6+x^{2}=\frac{7}{9}$$

Subtract 6 from both sides.

$$x^{2}=\frac{7}{9}-6$$

Subtract 6 from \frac{7}{9} to get -\frac{47}{9}.

$$x^{2}=-\frac{47}{9}$$

The equation is now solved.

$$x=\frac{\sqrt{47}i}{3} x=-\frac{\sqrt{47}i}{3}$$

Steps Using the Quadratic Formula

Rewrite the square root of the division \frac{49}{81} as the division of square roots \frac{\sqrt{49}}{\sqrt{81}}. Take the square root of both numerator and denominator.

$$6+x^{2}=\frac{7}{9}$$

Subtract \frac{7}{9} from both sides.

$$6+x^{2}-\frac{7}{9}=0$$

Subtract \frac{7}{9} from 6 to get \frac{47}{9}.

$$\frac{47}{9}+x^{2}=0$$

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.

$$x^{2}+\frac{47}{9}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 0 for b, and \frac{47}{9} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\times \frac{47}{9}}}{2}$$

Square 0.

$$x=\frac{0±\sqrt{-4\times \frac{47}{9}}}{2}$$

Multiply -4 times \frac{47}{9}.

$$x=\frac{0±\sqrt{-\frac{188}{9}}}{2}$$

Take the square root of -\frac{188}{9}.

$$x=\frac{0±\frac{2\sqrt{47}i}{3}}{2}$$

Now solve the equation x=\frac{0±\frac{2\sqrt{47}i}{3}}{2} when ± is plus.

$$x=\frac{\sqrt{47}i}{3}$$

Now solve the equation x=\frac{0±\frac{2\sqrt{47}i}{3}}{2} when ± is minus.

$$x=-\frac{\sqrt{47}i}{3}$$

The equation is now solved.

$$x=\frac{\sqrt{47}i}{3} x=-\frac{\sqrt{47}i}{3}$$