 # 2.6 Quadratic Equations, SPM Practice (Paper 2)

2.6 Quadratic Equations, SPM Practice (Paper 2)

Question 3:
If α and β are the roots of the quadratic equation 3x2 + 2x– 5 = 0, form the quadratic equations that have the following roots.

Solution:
3x2 + 2x – 5 = 0
a = 3, b = 2, c = –5
The roots are α and β.
$\begin{array}{l}\alpha +\beta =-\frac{b}{a}=-\frac{2}{3}\\ \alpha \beta =\frac{c}{a}=\frac{-5}{3}\end{array}$

(a)

Using the formula, x2– (sum of roots)x + product of roots = 0
The new quadratic equation is
${x}^{2}-\left(\frac{4}{5}\right)x+\left(-\frac{12}{5}\right)=0$
5x2 – 4x– 12 = 0

(b)
$\begin{array}{l}=\alpha +\beta +\left(\frac{2}{\alpha }+\frac{2}{\beta }\right)=\alpha +\beta +\frac{2\alpha +2\beta }{\alpha \beta }\\ =\alpha +\beta +\frac{2\left(\alpha +\beta \right)}{\alpha \beta }\\ =\frac{4}{5}+\frac{2\left(\frac{4}{5}\right)}{-\frac{12}{5}}\\ =\frac{4}{5}-\frac{2}{3}=\frac{2}{15}\end{array}$
$\begin{array}{l}=-\frac{12}{5}+4+\frac{4}{-\frac{12}{5}}\\ =-\frac{12}{5}+4-\frac{5}{3}=-\frac{1}{15}\end{array}$

The new quadratic equation is
${x}^{2}-\left(\frac{2}{15}\right)x+\left(-\frac{1}{15}\right)=0$
15x2 – 2x– 1 = 0

Question 4:
It is given α and β are the roots of the quadratic equation x (x – 3) = 2k – 4, where k is a constant.

Solution: