**2.6 Quadratic Equations, SPM Practice (Paper 2)**

**Question 1:**

(a) Find the values of

*k*if the equation (1 –*k*)*x*^{2}– 2(*k*+ 5)*x*+*k*+ 4 = 0 has real and equal roots.Hence, find the roots of the equation based on the values of

*k*obtained.(b) Given the curve

*y*= 5 + 4*x*–*x*^{2 }has tangen equation in the form*y*=*px*+ 9. Calculate the possible values of*p.*

*Solutions:***(a)**

For equal roots,

*b*

^{2 }– 4

*ac*= 0

[–2(

*k*+ 5)]^{ 2}– 4(1 –*k*)(*k*+ 4) = 04(

*k*+ 5)^{ 2 }– 4(1 –*k*)(*k*+ 4) = 04(

*k*^{2}+ 10*k*+ 25) – 4(4 – 3*k*–*k*^{2}) = 04

*k*^{2}+ 40*k*+ 100 – 16 + 12*k*+ 4*k*^{2}= 08

*k*^{2}+ 52*k*+ 84 = 02

*k*^{2}+ 13*k*+ 21 = 0(2

*k*+ 7) (*k*+ 3) = 0
$k=-\frac{7}{2},\text{}-3$
, −3

If
$k=-\frac{7}{2}$
, the equation is

$\begin{array}{l}\left(1+\frac{7}{2}\right){x}^{2}-2\left(-\frac{7}{2}+5\right)x-\frac{7}{2}+4=0\\ \frac{9}{2}{x}^{2}-3x+\frac{1}{2}=0\end{array}$

9

*x*

^{2}– 6

*x*+ 1 = 0

(3

*x*– 1) (3*x*– 1) = 0*x*= ⅓

If

*k*= –3, the equation is(1 + 3)

*x*^{ 2 }– 2(–3 + 5)*x*– 3 + 4 = 04

*x*^{2}– 4*x*+ 1 = 0(2

*x*– 1) (2*x*– 1) = 0*x*= ½

**(b)**

*y*= 5 + 4

*x*–

*x*

^{2 }----- (1)

*y*=

*px*+ 9 ---------- (2)

(1) = (2), 5 + 4

*x*–*x*^{2}=*px*+ 9*x*

^{2}+

*px*– 4

*x*+ 9 – 5 = 0

*x*

^{2}+ (

*p*– 4)

*x*+ 4 = 0

Tangen equation only has one intersection point with equal roots.

*b*

^{2 }– 4

*ac*= 0

(

*p*– 4)^{2}*– 4(1)(4) = 0**p*

^{2}– 8

*p*+ 16 – 16 = 0

*p*

^{2}– 8

*p*= 0

*p*(

*p*– 8) = 0

**Therefore,**

*p*= 0 and*p*= 8.