2.6 Quadratic Equations, SPM Practice (Paper 2)


2.6 Quadratic Equations, SPM Practice (Paper 2)
Question 1:
(a)  Find the values of k if the equation (1 – k) x2– 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 + 4x x2 has tangen equation in the form y = px + 9. Calculate the possible values of p.

Solutions:
(a)
For equal roots,
b2 – 4ac = 0
[–2(k + 5)] 2 – 4(1 – k)( k + 4) = 0
4(k + 5) 2 – 4(1 – k)( k + 4) = 0
4(k2 + 10k + 25) – 4(4 – 3k k2) = 0
4k2 + 40k + 100 – 16 + 12k + 4k2 = 0
8k2 + 52k + 84 = 0
2k2 + 13k + 21 = 0
(2k + 7) (k+ 3) = 0
k = 7 2 ,   3 , −3 

If k = 7 2  , the equation is
( 1 + 7 2 ) x 2 2 ( 7 2 + 5 ) x 7 2 + 4 = 0 9 2 x 2 3 x + 1 2 = 0  

9x2 – 6x + 1 = 0
(3x – 1) (3x – 1) = 0
x =

If k = –3, the equation is
(1 + 3)x 2 – 2(–3 + 5)x – 3 + 4 = 0
4x2 – 4x + 1 = 0
(2x – 1) (2x – 1) = 0
x = ½

(b)
y = 5 + 4x x2 ----- (1)
y = px + 9 ---------- (2)
(1)  = (2), 5 + 4x x2= px + 9
x2 + px – 4x + 9 – 5 = 0
x2 + (p – 4)x + 4 = 0

Tangen equation only has one intersection point with equal roots.
b2 – 4ac = 0
(p – 4)2 – 4(1)(4) = 0
p2 – 8p + 16 – 16 = 0
p2 – 8p = 0
p (p – 8) = 0
Therefore, p = 0 and p = 8.

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