Question

65.a insert a monomial into each trinomial such that the result is a perfect square. __ + 12x + 1 answer \boxed{} + 12x + 1

Explanation:

Step1: Recall perfect square formula

The perfect square trinomial formula is \((a + b)^2 = a^2 + 2ab + b^2\) or \((a - b)^2 = a^2 - 2ab + b^2\). Here we have \(2ab = 12x\) and \(b^2=1\), so \(b = \pm1\).

Step2: Solve for \(a\)

If \(b = 1\), from \(2ab=12x\), substitute \(b = 1\), we get \(2a\times1=12x\), so \(a = 6x\). Then \(a^2=(6x)^2 = 36x^2\).
If \(b=- 1\), from \(2ab = 12x\), substitute \(b=-1\), we get \(2a\times(-1)=12x\), so \(a=-6x\), and \(a^2=(-6x)^2 = 36x^2\) (same as above). Also, if we consider the form where the middle term is \(2ab\), and we can also think of the monomial as the square of the coefficient related to \(x\) or the constant term. But in the standard perfect square trinomial for the form \(a^2+2ab + b^2\), here \(2ab = 12x\) and \(b^2 = 1\), so \(a^2=36x^2\). Also, we can have other forms, but the most common is when we have the square of the linear term. Wait, also, if we consider the monomial as the square of the coefficient of \(x\) divided by 2? No, the formula is \((mx + n)^2=m^2x^2+2mnx + n^2\). So comparing with \(\_+12x + 1\), we have \(2mnx=12x\) and \(n^2 = 1\), so \(n=\pm1\). If \(n = 1\), then \(2m\times1=12\), so \(m = 6\), then \(m^2x^2=36x^2\). If \(n=- 1\), \(2m\times(-1)=12\), \(m=-6\), \(m^2x^2 = 36x^2\). Also, we can have the monomial as \(\frac{(12x)^2}{4\times1}=\frac{144x^2}{4}=36x^2\) (using the formula for completing the square: for \(ax^2+bx + c\) to be a perfect square, \(c=\frac{b^2}{4a}\), but here we are finding \(a\) when \(b = 12x\) and \(c = 1\), so \(a=\frac{b^2}{4c}=\frac{(12x)^2}{4\times1}=36x^2\)).

Answer:

\(36x^2\)