Quadratic Functions, SPM Practice (Long Questions)

Question 4:
(a) Find the range of values of k if the equation x² – kx + 3k – 5 = 0 does not have real roots.
(b) Show that the quadratic equation hx² – (h + 3)x + 1 = 0 has real and distinc roots for all values of h.

Solution:
(a)
$\begin{array}{l}{x}^2-kx+\left(3k-5\right)=0\\ \text{If the above equation has no real root,}\\ \therefore b^2-4ac<0.\\ k^2-4\left(3k-5\right)<0\\ k^2-12k+20<0\\ \left(k-2\right)\left(k-10\right)<0\end{array}$

Graph function y = (k – 2)(k – 10) cuts the horizontal line at k = 2 and k = 10 when b² – 4ac < 0.



The range of values of k that satisfy the inequality above is 2 < k < 10.

(b)
$\begin{array}{l}h{x}^2-\left(h+3\right)x+1=0\\ b^2-4ac={\left(h+3\right)}^2-4\left(h\right)\left(1\right)\\ =h^2+6h+9-4h\\ =h^2+2h+9\\ ={\left(h+\frac{2}{2}\right)}^2-{\left(\frac{2}{2}\right)}^2+9\\ ={\left(h+1\right)}^2-1+9\\ ={\left(h+1\right)}^2+8\end{array}$

The minimum value of (h + 1) + 8 is 8, a positive value. Therefore, b² – 4ac > 0 for all values of h.
Hence, quadratic equation hx² – (h + 3)x + 1 = 0 has real and distinc roots for all values of h.