If the roots of the equations ax² + 2bx + c = 0 and bx²-2√acx+b=0 are simultaneously real, then prove that b²= ac.
Answers
SOLUTION :
Given : ax² + 2bx + c = 0 …………(1)
and bx² - 2√acx + b = 0…………..(2)
On comparing the given equation with Ax² + Bx + C = 0
Let D1 & D2 be the discriminants of the two given equations .
For eq 1 :
Here, A = a , B = 2b , C = c
D(discriminant) = B² – 4AC
D1 = (2b)² - 4 × a × C
D1 = 4b² - 4ac ………(3)
For eq 2 :
bx² - 2√acx + b = 0
Here, A = b , B = - 2√ac, C = b
D(discriminant) = B² – 4AC
D2 = (- 2√ac)² - 4 × b × b
D2 = 4ac - 4b² …………(4)
Given : Roots are real for both the Given equations i.e D ≥ 0.
D1 ≥ 0
4b² - 4ac ≥ 0
[From eq 3]
4b² ≥ 4ac
b² ≥ ac ………….(5)
D2 ≥ 0
4ac - 4b² ≥ 0
4ac ≥ 4b²
ac ≥ b² …………(6)
From eq 5 & 6 ,
b² = ac
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Answer:
b² = ac
Step-by-step explanation:
(i) ax² + 2bx + c = 0
On comparing with ax² + bx + c = 0, we get
a = a, b = 2b, c = c
Given that the equation has real roots.
b² - 4ac ≥ 0
⇒ (2b)² - 4ac ≥ 0
⇒ 4b² - 4ac ≥ 0
⇒ b² - ac ≥ 0
⇒ b² ≥ ac
(ii) bx² - 2√acx + b² = 0
On comparing with ax² + bx + c = 0, we get
a = b, b = -2√ac, c = b
Given that the equation has real roots.
b² - 4ac ≥ 0
⇒ (-2√ac)² - 4(b)(b) ≥ 0
⇒ 4ac - 4b² ≥ 0
⇒ ac - b² ≥ 0
⇒ ac ≥ b²
From (i) & (ii), we get
b² = ac.
Hope it helps!