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

(62)^{2} = x^{2} + x^{2}

15 + 7 k = 2 k^{2}

2 x^{2} + 12 x + 2 = 0

4 p^{2} - 5 p - 6 = - 2

9 x^{2} - 11 = 5

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