What are the solutions to the equation r^2=(4)^2+(-3)^2 ?

The solutions to the equation r^2=(4)^2+(-3)^2 are r=5,r=-5

Find the zeros of r^2=(4)^2+(-3)^2

The zeros of r^2=(4)^2+(-3)^2 are r=5,r=-5

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r^{2} = (4)^{2} + (- 3)^{2}

r = 5, r = - 5

Solution steps

r^{2} = (4)^{2} + (- 3)^{2}

Simplify

4^{2} + (- 3)^{2} : 25

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

r = 25, r = - 25

25 = 5

- 25 = - 5

r = 5, r = - 5

3 X^{2} + 5 X - 2 = 0

2 x^{2} + 25 = - x^{2} + 10 x

(1 - y) (y - 1) = y^{2}

x^{2} - 5 = \frac{4 x ^{2}}{5}

3 x^{2} - 24 x - 36 = 0

What are the solutions to the equation r^2=(4)^2+(-3)^2 ?
The solutions to the equation r^2=(4)^2+(-3)^2 are r=5,r=-5
Find the zeros of r^2=(4)^2+(-3)^2
The zeros of r^2=(4)^2+(-3)^2 are r=5,r=-5