What are the solutions to the equation 10a(1+r+r^2)-35ar=0 ?

The solutions to the equation 10a(1+r+r^2)-35ar=0 are r=2,r= 1/2 ;\quad a\ne 0

Find the zeros of 10a(1+r+r^2)-35ar=0

The zeros of 10a(1+r+r^2)-35ar=0 are r=2,r= 1/2 ;\quad a\ne 0

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10 a (1 + r + r^{2}) - 35 a r = 0

r = 2, r = \frac{1}{2}; a \neq = 0

Decimal Notation

r = 2, r = 0.5

Solution steps

10 a (1 + r + r^{2}) - 35 a r = 0

Expand

10 a (1 + r + r^{2}) - 35 a r : 10 a r^{2} - 25 a r + 10 a

10 a r^{2} - 25 a r + 10 a = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- ( - 25 a ) \pm ( - 25 a ) ^{2} - 4 \cdot 10 a \cdot 10 a}{2 \cdot 10 a}

Simplify

(- 25 a)^{2} - 4 \cdot 10 a \cdot 10 a : 15 a

r_{1, 2} = \frac{- ( - 25 a ) \pm 15 a}{2 \cdot 10 a}; a \neq = 0

Separate the solutions

r_{1} = \frac{- ( - 25 a ) + 15 a}{2 \cdot 10 a}, r_{2} = \frac{- ( - 25 a ) - 15 a}{2 \cdot 10 a}

r = \frac{- ( - 25 a ) + 15 a}{2 \cdot 10 a} : 2

r = \frac{- ( - 25 a ) - 15 a}{2 \cdot 10 a} : \frac{1}{2}

The solutions to the quadratic equation are:

r = 2, r = \frac{1}{2}; a \neq = 0

(x + 2)^{2} + (x - 1) (x + 2) = 5 x

x^{2} + 6 x = 41

x^{2} - 4 x - 437 = 0

3 x (x + 2) = 2

x^{2} + 3 x - 72 = 0

What are the solutions to the equation 10a(1+r+r^2)-35ar=0 ?
The solutions to the equation 10a(1+r+r^2)-35ar=0 are r=2,r= 1/2 ;\quad a\ne 0
Find the zeros of 10a(1+r+r^2)-35ar=0
The zeros of 10a(1+r+r^2)-35ar=0 are r=2,r= 1/2 ;\quad a\ne 0