Question

Solve the equation for all values of x. -3(36x^2-25)(x^2+1)=0 (square root answer)

Answer 1

On solving the equation, we get the following values of x- 5/6, -5/6 and √-1.

If there exists an equation of the form (x-a)(x-b) = 0, then (x-a) and (x-b) are known as the factors of the equation and x = a and x =b are the roots of the equation.

Here, we are given -3 (36x^2 - 25) (x^2 + 1) = 0

We can find the value of x by equating each of the terms to 0

if (36x^2 - 25) = 0

⇒ 36x^2 = 25

x^2 = 25/36

x = sqrt(25/36)

x = 5/6 or x = -5/6

Similarly, if  (x^2 + 1) = 0

⇒ x^2 = - 1

⇒ x = sqrt(- 1)

This will come out to be an imaginary number, since square root of negative numbers are not real.  √

Thus, the values of x are-  5/6, -5/6 and √-1.

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