What are the solutions to the equation (36-R)^2+(18-R)^2=R^2 ?

The solutions to the equation (36-R)^2+(18-R)^2=R^2 are R=90,R=18

Find the zeros of (36-R)^2+(18-R)^2=R^2

The zeros of (36-R)^2+(18-R)^2=R^2 are R=90,R=18

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(36 - R)^{2} + (18 - R)^{2} = R^{2}

R = 90, R = 18

Solution steps

(36 - R)^{2} + (18 - R)^{2} = R^{2}

Expand

(36 - R)^{2} + (18 - R)^{2} : 2 R^{2} - 108 R + 1620

2 R^{2} - 108 R + 1620 = R^{2}

Move

R^{2}

to the left side

R^{2} - 108 R + 1620 = 0

Solve with the quadratic formula

R_{1, 2} = \frac{- ( - 108 ) \pm ( - 108 ) ^{2} - 4 \cdot 1 \cdot 1620}{2 \cdot 1}

(- 108)^{2} - 4 \cdot 1 \cdot 1620 = 72

R_{1, 2} = \frac{- ( - 108 ) \pm 72}{2 \cdot 1}

Separate the solutions

R_{1} = \frac{- ( - 108 ) + 72}{2 \cdot 1}, R_{2} = \frac{- ( - 108 ) - 72}{2 \cdot 1}

R = \frac{- ( - 108 ) + 72}{2 \cdot 1} : 90

R = \frac{- ( - 108 ) - 72}{2 \cdot 1} : 18

The solutions to the quadratic equation are:

R = 90, R = 18

5 x^{2} - 20 x - 12 = 0

4 x^{2} + x - \frac{1}{16} = 0

9 x^{2} - 6 x + 1 = 0

(x - 2)^{2} - 5 = 0

(2 w + 1) w = 66

What are the solutions to the equation (36-R)^2+(18-R)^2=R^2 ?
The solutions to the equation (36-R)^2+(18-R)^2=R^2 are R=90,R=18
Find the zeros of (36-R)^2+(18-R)^2=R^2
The zeros of (36-R)^2+(18-R)^2=R^2 are R=90,R=18