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=−72, −3
If
k=−72
, the equation is
(1+72)x2−2(−72+5)x−72+4=092x2−3x+12=0
9x2 – 6x + 1 = 0
(3x – 1) (3x – 1) = 0
x = ⅓
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.
Question 2:
Given α and β are two roots of the quadratic equation (2x + 5)(x + 1) + p = 0 where αβ = 3 and p is a constant.
Find the value p, α and of β.
Solutions:
(2x + 5)(x + 1) + p = 0
2x2 + 2x + 5x + 5 + p = 0
2x2 + 7x + 5 + p = 0
*Compare with, x2– (sum of roots)x + product of roots = 0
x2+72x+5+p2=0←divide both sides with 2
Product of roots, αβ = 3
5+p2=3
5 + p = 6
p = 1
Sum of roots =
−72
α+β=−72 → (1)and αβ=3 → (2)from (2), β=3α → (3)Substitute (3) into (1),α+3α=−72
2α2+ 6 = –7α ← (multiply both sides with 2α)
2α2+ 7α + 6 = 0
(2α + 3)(α + 2) = 0
2α + 3 = 0 or α + 2 = 0
α=− 3 2 α = –2
Substitute α=−32 into (3),β=3−32=3(−23)=−2
Substitute α = –2 into (3),
β=−32∴ p=1, andwhen α=−32,β=−2 and α=−2,β=−32.