Question

given $9x^4 - 6x^2 + 1 = (ax^2 - 1)^2$, determine the value of $a$.

Explanation:

Step1: Expand the right-hand side

We know that \((m - n)^2 = m^2 - 2mn + n^2\). Here, \(m = ax^2\) and \(n = 1\), so \((ax^2 - 1)^2=(ax^2)^2-2(ax^2)(1)+1^2=a^2x^4 - 2ax^2+1\).

Step2: Compare coefficients

The left-hand side is \(9x^4 - 6x^2 + 1\) and the right-hand side after expansion is \(a^2x^4 - 2ax^2+1\). We compare the coefficients of \(x^4\):
For the coefficient of \(x^4\), we have \(a^2 = 9\) (since the coefficient of \(x^4\) on the left is 9 and on the right is \(a^2\)). Also, we can check the coefficient of \(x^2\): \(- 2a=-6\) (coefficient of \(x^2\) on left is - 6 and on right is - 2a).
From \(-2a=-6\), we can solve for \(a\) by dividing both sides by - 2: \(a=\frac{- 6}{-2}=3\). We can also check with \(a^2 = 9\), the solutions for \(a\) from \(a^2 = 9\) are \(a = 3\) or \(a=-3\), but from the coefficient of \(x^2\) we see that \(a = 3\) (because if \(a=-3\), then \(-2a = 6\) which does not match the coefficient of \(x^2\) on the left which is - 6).

Answer:

3