What are the solutions to the equation 35(p^2-1)=24p ?

The solutions to the equation 35(p^2-1)=24p are p= 7/5 ,p=-5/7

Find the zeros of 35(p^2-1)=24p

The zeros of 35(p^2-1)=24p are p= 7/5 ,p=-5/7

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35 (p^{2} - 1) = 24 p

p = \frac{7}{5}, p = - \frac{5}{7}

Decimal Notation

p = 1.4, p = - 0.71428 \dots

Solution steps

35 (p^{2} - 1) = 24 p

Expand

35 (p^{2} - 1) : 35 p^{2} - 35

35 p^{2} - 35 = 24 p

Move

24 p

to the left side

35 p^{2} - 35 - 24 p = 0

Write in the standard form

a x^{2} + b x + c = 0

35 p^{2} - 24 p - 35 = 0

Solve with the quadratic formula

p_{1, 2} = \frac{- ( - 24 ) \pm ( - 24 ) ^{2} - 4 \cdot 35 ( - 35 )}{2 \cdot 35}

(- 24)^{2} - 4 \cdot 35 (- 35) = 74

p_{1, 2} = \frac{- ( - 24 ) \pm 74}{2 \cdot 35}

Separate the solutions

p_{1} = \frac{- ( - 24 ) + 74}{2 \cdot 35}, p_{2} = \frac{- ( - 24 ) - 74}{2 \cdot 35}

p = \frac{- ( - 24 ) + 74}{2 \cdot 35} : \frac{7}{5}

p = \frac{- ( - 24 ) - 74}{2 \cdot 35} : - \frac{5}{7}

The solutions to the quadratic equation are:

p = \frac{7}{5}, p = - \frac{5}{7}

(2 x + 4) (3 x - 2) = 0

\frac{8 x}{14 y} \cdot \frac{5 y ^{4}}{5 x ^{3}}

- 10 + 4 x \leq 10 x - 4

1 = \frac{1}{3} (10 x - 3)

What are the solutions to the equation 35(p^2-1)=24p ?
The solutions to the equation 35(p^2-1)=24p are p= 7/5 ,p=-5/7
Find the zeros of 35(p^2-1)=24p
The zeros of 35(p^2-1)=24p are p= 7/5 ,p=-5/7