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# Which quadratic equation is equivalent to (x^{2} - 1)^{2} - 11(x^{2} - 1) + 24 = 0?

**Solution:**

The question is based on the standard form of a quadratic equation.

Let's write the equation in it's standard form.

(x^{2} - 1)^{2} - 11(x^{2} - 1) + 24 = 0

Consider (x^{2} - 1) as u

u^{2 }- 11u + 24 = 0

u^{2 }- 8u - 3u + 24 = 0

u(u - 8) - 3( u - 8) = 0 → by splitting the middle term

(u - 8) ( u - 3) = 0

(x^{2} - 1 - 8 )( x^{2} - 1 - 3) = 0 ---------> by putting (x^{2} - 1) = u

(x^{2} - 9 )( x^{2} - 4) = 0

Hence, x^{2} – 9 = 0 or x^{2} – 4 = 0 -----> [ similar to the quadratic equation ax^{2} + bx + c = 0 ]

Thus, (x^{2} - 9) = 0 or (x^{2} - 4) = 0 are the standard form of (x^{2} - 1)^{2} - 11(x^{2} - 1) + 24 = 0.

## Which quadratic equation is equivalent to (x^{2} - 1)^{2} - 11(x^{2} - 1) + 24 = 0?

**Summary:**

(x^{2} - 9) = 0 or (x^{2} - 4) = 0 are the standard equivalent form to (x^{2} - 1)^{2} - 11(x^{2} - 1) + 24 = 0.

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