Solve sqrt{a^2-4a+20}=a

$$\sqrt { a ^ { 2 } - 4 a + 20 } = a$$

Solve for a

$$a=5$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{a^{2}-4a+20}\right)^{2}=a^{2}$$

Calculate \sqrt{a^{2}-4a+20} to the power of 2 and get a^{2}-4a+20.

$$a^{2}-4a+20=a^{2}$$

Subtract a^{2} from both sides.

$$a^{2}-4a+20-a^{2}=0$$

Combine a^{2} and -a^{2} to get 0.

$$-4a+20=0$$

Subtract 20 from both sides. Anything subtracted from zero gives its negation.

$$-4a=-20$$

Divide both sides by -4.

$$a=\frac{-20}{-4}$$

Divide -20 by -4 to get 5.

$$a=5$$

Substitute 5 for a in the equation \sqrt{a^{2}-4a+20}=a.

$$\sqrt{5^{2}-4\times 5+20}=5$$

Simplify. The value a=5 satisfies the equation.

$$5=5$$

Equation \sqrt{a^{2}-4a+20}=a has a unique solution.

$$a=5$$