What are the solutions to the equation 63(p^2-1)=32p ?

The solutions to the equation 63(p^2-1)=32p are p= 9/7 ,p=-7/9

Find the zeros of 63(p^2-1)=32p

The zeros of 63(p^2-1)=32p are p= 9/7 ,p=-7/9

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63 (p^{2} - 1) = 32 p

p = \frac{9}{7}, p = - \frac{7}{9}

Decimal Notation

p = 1.28571 \dots, p = - 0.77777 \dots

Solution steps

63 (p^{2} - 1) = 32 p

Expand

63 (p^{2} - 1) : 63 p^{2} - 63

63 p^{2} - 63 = 32 p

Move

32 p

to the left side

63 p^{2} - 63 - 32 p = 0

Write in the standard form

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

63 p^{2} - 32 p - 63 = 0

Solve with the quadratic formula

p_{1, 2} = \frac{- ( - 32 ) \pm ( - 32 ) ^{2} - 4 \cdot 63 ( - 63 )}{2 \cdot 63}

(- 32)^{2} - 4 \cdot 63 (- 63) = 130

p_{1, 2} = \frac{- ( - 32 ) \pm 130}{2 \cdot 63}

Separate the solutions

p_{1} = \frac{- ( - 32 ) + 130}{2 \cdot 63}, p_{2} = \frac{- ( - 32 ) - 130}{2 \cdot 63}

p = \frac{- ( - 32 ) + 130}{2 \cdot 63} : \frac{9}{7}

p = \frac{- ( - 32 ) - 130}{2 \cdot 63} : - \frac{7}{9}

The solutions to the quadratic equation are:

p = \frac{9}{7}, p = - \frac{7}{9}

- 5 x^{2} - 67 x - 100 = - 7 x

x^{2} + 3 x + 6

a x^{2} + (- 2 a^{2} + 2) x - 4 a = 0

5 x^{2} = 46 + 9

8 m^{2} = 15

What are the solutions to the equation 63(p^2-1)=32p ?
The solutions to the equation 63(p^2-1)=32p are p= 9/7 ,p=-7/9
Find the zeros of 63(p^2-1)=32p
The zeros of 63(p^2-1)=32p are p= 9/7 ,p=-7/9