Find the value of p for which $x^{2} - x + p$ becomes a perfect square.

Assessment: WAEC SSCE - Further Mathematics - 2007 (Objective) Subject: Further Mathematics

Question 1 Report

$-\frac{1}{2}$

$\frac{1}{4}$

$\frac{1}{2}$

$1$

Answer Details

To make the expression $x^2-x+p$ a perfect square, we need to add and subtract a suitable constant term. To do this, we can start by finding the square of half the coefficient of the linear term. That is, we find the square of $\frac{1}{2}$ which is $\frac{1}{4}$. So, we have: $$x^2-x+\frac{1}{4}-\frac{1}{4}+p$$ Now, we can write this expression as a perfect square by factoring the first three terms: $$\left(x-\frac{1}{2}\right)^2+p-\frac{1}{4}$$ For this expression to be a perfect square, the last term must be zero, so we have: $$p-\frac{1}{4}=0$$ Solving for $p$, we get: $$p=\frac{1}{4}$$ Therefore, the value of $p$ that makes $x^2-x+p$ a perfect square is $\frac{1}{4}$.

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Find the value of p for which \(x^{2} - x + p\) becomes a perfect square.

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Find the value of p for which \(x^{2} - x + p\) becomes a perfect square.

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