# SPM Questions for Quadratic Equations

We will discuss some pass year SPM exam questions based on:

• Solve the quadratic equations – SPM 2003, SPM 2005
• Form a quadratic equation – SPM 2009
• Determine the conditions for the type of roots – SPM2010

### Solve the quadratic equations – SPM 2003 Paper 1 Question 3

(3 marks)  $\begin{array}{rcl} 2x(x-4) & = & (1-x)(x+2) \\ 2{{x}^{2}}-8x &=& x+2-{{x}^{2}}-2x \\ 2{{x}^{2}}+{{x}^{2}}-8x-x+2x &=& 2 \\ 3{{x}^{2}}-7x &=& 2 \\ {{x}^{2}}-\frac{7}{3}x &=& \frac{2}{3} \\ {{x}^{2}}-\frac{7}{3}x+{{\left( -\frac{7}{3(2)} \right)}^{2}} &=& \frac{2}{3}+{{\left( -\frac{7}{3(2)} \right)}^{2}} \\ {{x}^{2}}-\frac{7}{3}x+{{\left( -\frac{7}{6} \right)}^{2}} &=& \frac{2}{3}+{{\left( -\frac{7}{6} \right)}^{2}} \\ {{\left( x-\frac{7}{6} \right)}^{2}} &=& \frac{73}{36} \\ x-\frac{7}{6} &=& \pm \sqrt{\frac{73}{36}} \\ x &=& \frac{7}{6}+\sqrt{\frac{73}{36}}\text \quad {or} \quad x = \frac{7}{6}-\sqrt{\frac{73}{36}} \\ x &=& \underline {2.59} \text \qquad {or} \quad x = \underline {- 0.2573} \\ \end{array}$

Note: If you have any doubt, please refer back my previous post on how to solve the equation using completing the square method.

### SPM 2005 Paper1 Question 5

(3 marks)  $\begin{array}{rcl} x(2x-5)&=&2x-1 \\ 2{{x}^{2}}-5x-2x+1&=&0 \\ 2{{x}^{2}}-7x+1&=&0 \\ hence, a=2,b=-7,c&=&1 \\ x&=& \frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} \\ x&=& \frac{-(-7)\pm \sqrt{{{(-7)}^{2}}-4(2)(1)}}{2(2)} \\ x&=& \frac{7+\sqrt{41}}{4}\text\quad{ or }\quad x=\frac{7-\sqrt{41}}{4} \\ x&=& \underline{3.351}\text\quad{ or }\quad x=\underline{0.149} \\ \end{array}$

### Form a quadratic equation – SPM 2009 Paper 2 Question 2

The quadratic equation x2 – 5x + 6 = 0 has roots h and k, where h > k.

(a) Find

i) the value of h and of k.
ii) the range of x if x2 – 5x + 6 > 0
(5 marks)

(b) Using the values of h and k from 2(a)(i), form the quadratic equation which has roots h + 2 and 3k – 2.
(2 marks)

a i) For x– 5x + 6 = 0, given roots are h and k, ii) (b) $h + 2$ $= 3 + 2$ $= 5$ $3k - 2$ $= 3(2) - 2$ $= 4$ $SOR = 5 + 4 = 9$ $\text{POR = 5 x 4 = 20}$ $Equations:$ ${{x}^{2}}-(\text{SOR)}x+(\text{POR)}=0$ $\underline{{{x}^{2}}-9x+20=0}$

### Determine the conditions for the type of roots – SPM 2010 Paper1 Question 5

The quadratic equation (1-p)x2 – 6x +10 = 0, where p is a constant, has two different roots. Find the range of values of p.

(3 marks)  $\begin{array}{rcl} (1-p){{x}^{2}}-6x+10=0 \\ a=1-p,b=-6,c=10 \\ \\ \text{2 different roots, meant } \\ {{b}^{2}}-4ac>0 \\ {{(-6)}^{2}}-4(1-p)(10)>0 \\ 36-40+40p>0 \\ 40p>4 \\ p>\frac{4}{40} \\ \underline{p>\frac{1}{10}} \\ \end{array}$ 