Solve 9^2+b^2=41^2

$$9 ^ { 2 } + b ^ { 2 } = 41 ^ { 2 }$$

Solve for b

$$b=-40$$

$$b=40$$

Steps Using Difference of Squares

Calculate 9 to the power of 2 and get 81.

$$81+b^{2}=41^{2}$$

Calculate 41 to the power of 2 and get 1681.

$$81+b^{2}=1681$$

Subtract 1681 from both sides.

$$81+b^{2}-1681=0$$

Subtract 1681 from 81 to get -1600.

$$-1600+b^{2}=0$$

Consider -1600+b^{2}. Rewrite -1600+b^{2} as b^{2}-40^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(b-40\right)\left(b+40\right)=0$$

To find equation solutions, solve b-40=0 and b+40=0.

$$b=40 b=-40$$

Steps by Finding Square Root

Calculate 9 to the power of 2 and get 81.

$$81+b^{2}=41^{2}$$

Calculate 41 to the power of 2 and get 1681.

$$81+b^{2}=1681$$

Subtract 81 from both sides.

$$b^{2}=1681-81$$

Subtract 81 from 1681 to get 1600.

$$b^{2}=1600$$

Take the square root of both sides of the equation.

$$b=40 b=-40$$

Steps Using the Quadratic Formula

Calculate 9 to the power of 2 and get 81.

$$81+b^{2}=41^{2}$$

Calculate 41 to the power of 2 and get 1681.

$$81+b^{2}=1681$$

Subtract 1681 from both sides.

$$81+b^{2}-1681=0$$

Subtract 1681 from 81 to get -1600.

$$-1600+b^{2}=0$$

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.

$$b^{2}-1600=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 0 for b, and -1600 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$b=\frac{0±\sqrt{0^{2}-4\left(-1600\right)}}{2}$$

Square 0.

$$b=\frac{0±\sqrt{-4\left(-1600\right)}}{2}$$

Multiply -4 times -1600.

$$b=\frac{0±\sqrt{6400}}{2}$$

Take the square root of 6400.

$$b=\frac{0±80}{2}$$

Now solve the equation b=\frac{0±80}{2} when ± is plus. Divide 80 by 2.

$$b=40$$

Now solve the equation b=\frac{0±80}{2} when ± is minus. Divide -80 by 2.

$$b=-40$$

The equation is now solved.

$$b=40 b=-40$$