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

3 t^{2} - 10 t - 6 = 2

9 y^{2} + 1.2 y

5 x^{2} - 32 x - 21 = 0

x^{2} - 2 x + 8 = 2

6 x^{2} - 3 = 8 x

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