What are the solutions to the equation (18-R)^2+(9-R)^2=R^2 ?

The solutions to the equation (18-R)^2+(9-R)^2=R^2 are R=45,R=9

Find the zeros of (18-R)^2+(9-R)^2=R^2

The zeros of (18-R)^2+(9-R)^2=R^2 are R=45,R=9

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(18 - R)^{2} + (9 - R)^{2} = R^{2}

R = 45, R = 9

Solution steps

(18 - R)^{2} + (9 - R)^{2} = R^{2}

Expand

(18 - R)^{2} + (9 - R)^{2} : 2 R^{2} - 54 R + 405

2 R^{2} - 54 R + 405 = R^{2}

Move

R^{2}

to the left side

R^{2} - 54 R + 405 = 0

Solve with the quadratic formula

R_{1, 2} = \frac{- ( - 54 ) \pm ( - 54 ) ^{2} - 4 \cdot 1 \cdot 405}{2 \cdot 1}

(- 54)^{2} - 4 \cdot 1 \cdot 405 = 36

R_{1, 2} = \frac{- ( - 54 ) \pm 36}{2 \cdot 1}

Separate the solutions

R_{1} = \frac{- ( - 54 ) + 36}{2 \cdot 1}, R_{2} = \frac{- ( - 54 ) - 36}{2 \cdot 1}

R = \frac{- ( - 54 ) + 36}{2 \cdot 1} : 45

R = \frac{- ( - 54 ) - 36}{2 \cdot 1} : 9

The solutions to the quadratic equation are:

R = 45, R = 9

3 x^{2} - 10 + 3 = 0

2 n^{2} - 5 n - 12 = 0

0 = x^{2} - 26 x + 48

(4 x^{2} - 6 x + 9) = 0

m^{2} - 256 = 0

What are the solutions to the equation (18-R)^2+(9-R)^2=R^2 ?
The solutions to the equation (18-R)^2+(9-R)^2=R^2 are R=45,R=9
Find the zeros of (18-R)^2+(9-R)^2=R^2
The zeros of (18-R)^2+(9-R)^2=R^2 are R=45,R=9