What are the solutions to the equation 5(x^2-1)=4(x^2+1) ?

The solutions to the equation 5(x^2-1)=4(x^2+1) are x=3,x=-3

Find the zeros of 5(x^2-1)=4(x^2+1)

The zeros of 5(x^2-1)=4(x^2+1) are x=3,x=-3

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5 (x^{2} - 1) = 4 (x^{2} + 1)

x = 3, x = - 3

Solution steps

5 (x^{2} - 1) = 4 (x^{2} + 1)

Expand

5 (x^{2} - 1) : 5 x^{2} - 5

Expand

4 (x^{2} + 1) : 4 x^{2} + 4

5 x^{2} - 5 = 4 x^{2} + 4

Move

5

to the right side

5 x^{2} = 4 x^{2} + 9

Move

4 x^{2}

to the left side

x^{2} = 9

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

x = 9, x = - 9

9 = 3

- 9 = - 3

x = 3, x = - 3

48 x^{2} + 12 x - 90 = 0

x^{2} - 4 x + 2 = 10

x, 4 x^{2} + 4 y^{2} = 0

a, x^{2} + 2 a x + a^{2} = \frac{( x \cdot y )}{p}

2 (x - 3) (x + 10) = 0

What are the solutions to the equation 5(x^2-1)=4(x^2+1) ?
The solutions to the equation 5(x^2-1)=4(x^2+1) are x=3,x=-3
Find the zeros of 5(x^2-1)=4(x^2+1)
The zeros of 5(x^2-1)=4(x^2+1) are x=3,x=-3