(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)\...
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 Details
(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,
\[k\approx3.8.\]
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