Solve 9-x^2/x-3=0quadx

$$\frac { 9 - x ^ { 2 } } { x - 3 } = 0 \quad x$$

Solve for x

$$x=-3$$

Steps by Finding Square Root

Variable x cannot be equal to 3 since division by zero is not defined. Multiply both sides of the equation by x-3.

$$9-x^{2}=0x\left(x-3\right)$$

Anything times zero gives zero.

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

Subtract 9 from both sides. Anything subtracted from zero gives its negation.

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

Divide both sides by -1.

$$x^{2}=\frac{-9}{-1}$$

Fraction \frac{-9}{-1} can be simplified to 9 by removing the negative sign from both the numerator and the denominator.

$$x^{2}=9$$

Take the square root of both sides of the equation.

$$x=3 x=-3$$

Variable x cannot be equal to 3.

$$x=-3$$

Steps Using the Quadratic Formula

Variable x cannot be equal to 3 since division by zero is not defined. Multiply both sides of the equation by x-3.

$$9-x^{2}=0x\left(x-3\right)$$

Anything times zero gives zero.

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

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.

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

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

$$x=\frac{0±\sqrt{0^{2}-4\left(-1\right)\times 9}}{2\left(-1\right)}$$

Square 0.

$$x=\frac{0±\sqrt{-4\left(-1\right)\times 9}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{0±\sqrt{4\times 9}}{2\left(-1\right)}$$

Multiply 4 times 9.

$$x=\frac{0±\sqrt{36}}{2\left(-1\right)}$$

Take the square root of 36.

$$x=\frac{0±6}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{0±6}{-2}$$

Now solve the equation x=\frac{0±6}{-2} when ± is plus. Divide 6 by -2.

$$x=-3$$

Now solve the equation x=\frac{0±6}{-2} when ± is minus. Divide -6 by -2.

$$x=3$$

The equation is now solved.

$$x=-3 x=3$$

Variable x cannot be equal to 3.

$$x=-3$$