What are the solutions to the equation (6-R)^2+(3-R)^2=R^2 ?

The solutions to the equation (6-R)^2+(3-R)^2=R^2 are R=15,R=3

Find the zeros of (6-R)^2+(3-R)^2=R^2

The zeros of (6-R)^2+(3-R)^2=R^2 are R=15,R=3

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(6 - R)^{2} + (3 - R)^{2} = R^{2}

R = 15, R = 3

Solution steps

(6 - R)^{2} + (3 - R)^{2} = R^{2}

Expand

(6 - R)^{2} + (3 - R)^{2} : 2 R^{2} - 18 R + 45

2 R^{2} - 18 R + 45 = R^{2}

Move

R^{2}

to the left side

R^{2} - 18 R + 45 = 0

Solve with the quadratic formula

R_{1, 2} = \frac{- ( - 18 ) \pm ( - 18 ) ^{2} - 4 \cdot 1 \cdot 45}{2 \cdot 1}

(- 18)^{2} - 4 \cdot 1 \cdot 45 = 12

R_{1, 2} = \frac{- ( - 18 ) \pm 12}{2 \cdot 1}

Separate the solutions

R_{1} = \frac{- ( - 18 ) + 12}{2 \cdot 1}, R_{2} = \frac{- ( - 18 ) - 12}{2 \cdot 1}

R = \frac{- ( - 18 ) + 12}{2 \cdot 1} : 15

R = \frac{- ( - 18 ) - 12}{2 \cdot 1} : 3

The solutions to the quadratic equation are:

R = 15, R = 3

x^{2} + 11 x - 30 = 0

n = - h^{2} + \frac{7}{a}

- 15 x^{2} + x + 28 = 0

(a - 5)^{2} = 2 a

3 x^{2} + 12 x = 7 x + 2

What are the solutions to the equation (6-R)^2+(3-R)^2=R^2 ?
The solutions to the equation (6-R)^2+(3-R)^2=R^2 are R=15,R=3
Find the zeros of (6-R)^2+(3-R)^2=R^2
The zeros of (6-R)^2+(3-R)^2=R^2 are R=15,R=3