(a) If \(\alpha\) and \(\beta\) are the roots of... | WAEC SSCE - Further Mathematics - 2008 (Essay)

Question 1 Report

(a) If \(\alpha\) and \(\beta\) are the roots of the equation \(2x^{2} + 5x - 6 = 0\), find the equation whose roots are \((\alpha - 2)\) and \((\beta - 2)\).

(b) Given that \(\int_{0} ^{k} (x^{2} - 2x) \mathrm {d} x = 4\), find the values of k.

Answer

(a) New equation whose roots are \((\alpha-2)\) and \((\beta-2).\)

For \(2x^2+5x-6=0,\) sum and product of roots:

\[\alpha+\beta=-\frac{5}{2},\qquad \alpha\beta=\frac{-6}{2}=-3.\]

Sum of new roots:

\[(\alpha-2)+(\beta-2)=(\alpha+\beta)-4=-\frac{5}{2}-4=-\frac{13}{2}.\]

Product of new roots:

\[(\alpha-2)(\beta-2)=\alpha\beta-2(\alpha+\beta)+4=-3-2\!\left(-\frac{5}{2}\right)+4=-3+5+4=6.\]

Required equation \(x^2-(\text{sum})x+(\text{product})=0:\)

\[x^2+\frac{13}{2}x+6=0\ \Rightarrow\ 2x^2+13x+12=0.\]

(b) Solve \(\displaystyle\int_0^k(x^2-2x)\,dx=4.\)

\[\int_0^k(x^2-2x)\,dx=\left[\frac{x^3}{3}-x^2\right]_0^k=\frac{k^3}{3}-k^2.\]

Set equal to 4 and multiply through by 3:

\[\frac{k^3}{3}-k^2=4\ \Rightarrow\ k^3-3k^2-12=0.\]

This cubic has a single real root. Testing shows \(f(3)=27-27-12=-12<0\) and \(f(4)=64-48-12=4>0,\) so the root lies between 3 and 4. Solving numerically,

\[k\approx3.8.\]

Read lesson note on Dynamics (WAEC)

Read lesson note on Polynomial Functions (WAEC)

Answer Details