What are the solutions to the equation (8-R)^2+(4-R)^2=R^2 ?

The solutions to the equation (8-R)^2+(4-R)^2=R^2 are R=20,R=4

Find the zeros of (8-R)^2+(4-R)^2=R^2

The zeros of (8-R)^2+(4-R)^2=R^2 are R=20,R=4

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(8 - R)^{2} + (4 - R)^{2} = R^{2}

R = 20, R = 4

Solution steps

(8 - R)^{2} + (4 - R)^{2} = R^{2}

Expand

(8 - R)^{2} + (4 - R)^{2} : 2 R^{2} - 24 R + 80

2 R^{2} - 24 R + 80 = R^{2}

Move

R^{2}

to the left side

R^{2} - 24 R + 80 = 0

Solve with the quadratic formula

R_{1, 2} = \frac{- ( - 24 ) \pm ( - 24 ) ^{2} - 4 \cdot 1 \cdot 80}{2 \cdot 1}

(- 24)^{2} - 4 \cdot 1 \cdot 80 = 16

R_{1, 2} = \frac{- ( - 24 ) \pm 16}{2 \cdot 1}

Separate the solutions

R_{1} = \frac{- ( - 24 ) + 16}{2 \cdot 1}, R_{2} = \frac{- ( - 24 ) - 16}{2 \cdot 1}

R = \frac{- ( - 24 ) + 16}{2 \cdot 1} : 20

R = \frac{- ( - 24 ) - 16}{2 \cdot 1} : 4

The solutions to the quadratic equation are:

R = 20, R = 4

(x) = 4 x^{2} - 3

0.25 = x^{2}

4 p (p - 2) = 4 - 2 p

x, x^{2} - 2 x - 48 = 0

x^{2} - 10 = 24

What are the solutions to the equation (8-R)^2+(4-R)^2=R^2 ?
The solutions to the equation (8-R)^2+(4-R)^2=R^2 are R=20,R=4
Find the zeros of (8-R)^2+(4-R)^2=R^2
The zeros of (8-R)^2+(4-R)^2=R^2 are R=20,R=4