Quadratic Equations, SPM Practice (Paper 2)

Posted on April 18, 2020 by user

2.10.2 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.

$\begin{array}{l} (a) \frac{2}{α} and \frac{2}{β} \\ (b) (α + \frac{2}{β}) and (β + \frac{2}{α}) \end{array}$

Solution:

3x² + 2x – 5 = 0

a = 3, b = 2, c = –5

The roots are α and β.

$\begin{array}{l} α + β = - \frac{b}{a} = - \frac{2}{3} \\ α β = \frac{c}{a} = \frac{- 5}{3} \end{array}$

(a)

$\begin{array}{l} The new roots are \frac{2}{α} and \frac{2}{β} . \\ Sum of new roots \\ = \frac{2}{α} + \frac{2}{β} = \frac{2 β + 2 α}{α β} \\ = \frac{2 (α + β)}{α β} = \frac{2 (- \frac{2}{3})}{- \frac{5}{3}} = \frac{4}{5} \end{array}$

$\begin{array}{l} Product of new roots \\ = (\frac{2}{α}) (\frac{2}{β}) = \frac{4}{α β} \\ = \frac{4}{- \frac{5}{3}} = - \frac{12}{5} \end{array}$

Using the formula, x²– (sum of roots)x + product of roots = 0

The new quadratic equation is

$x^{2} - (\frac{4}{5}) x + (- \frac{12}{5}) = 0$

5x² – 4x– 12 = 0

(b)

$\begin{array}{l} The new roots are (α + \frac{2}{β}) and (β + \frac{2}{α}) . \\ Sum of new roots \\ = (α + \frac{2}{β}) + (β + \frac{2}{α}) \end{array}$

$\begin{array}{l} = α + β + (\frac{2}{α} + \frac{2}{β}) = α + β + \frac{2 α + 2 β}{α β} \\ = α + β + \frac{2 (α + β)}{α β} \\ = \frac{4}{5} + \frac{2 (\frac{4}{5})}{- \frac{12}{5}} \\ = \frac{4}{5} - \frac{2}{3} = \frac{2}{15} \end{array}$

$\begin{array}{l} Product of new roots \\ = (α + \frac{2}{β}) (β + \frac{2}{α}) \\ = α β + 2 + 2 + \frac{4}{α β} \end{array}$

$\begin{array}{l} = - \frac{12}{5} + 4 + \frac{4}{- \frac{12}{5}} \\ = - \frac{12}{5} + 4 - \frac{5}{3} = - \frac{1}{15} \end{array}$

The new quadratic equation is

$x^{2} - (\frac{2}{15}) x + (- \frac{1}{15}) = 0$

15x² – 2x– 1 = 0