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WAEC SSCE - Further Mathematics - 2016 (Objective)

Question 1 Report

If the solution set of \(x^{2} + kx - 5 = 0\) is (-1, 5), find the value of k.

Answer Details
To solve this problem, we can use the fact that the sum and product of the roots of a quadratic equation are related to its coefficients. Specifically, for the equation \(ax^2+bx+c=0\), the sum of the roots is given by \(-b/a\) and the product of the roots is given by \(c/a\). In this problem, we are given that the solution set of \(x^2+kx-5=0\) is (-1, 5). This means that the quadratic equation has roots -1 and 5. Therefore, we can write the equation in factored form as \((x+1)(x-5)=0\). Expanding this expression, we get: $$x^2-4x-5=0$$ Comparing this to the original equation \(x^2+kx-5=0\), we can see that \(k=-4\). Therefore, the answer is (B) -4. Alternatively, we can use the sum and product of roots formula to solve for k. Since the roots of the equation are -1 and 5, we know that: $$-b/a=-1+5=4$$ and $$c/a=-5$$ Using the fact that \(a=1\), we can solve for \(b\) and \(c\): $$b=-a(-b/a)=4(1)=4$$ $$c=a(c/a)=1(-5)=-5$$ Therefore, the original equation is \(x^2+4x-5=0\), and the answer is (B) -4.