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If the sum of the roots of ax2 + bx + c = 0 is equal to the sum of the squares of their reciprocals, then which one of the following relation is correct?
- a)ab2 + bc2 = 2a2c
- b)ac2 + bc2 = 2b2a
- c)ab2 + bc2 = a2c
- d)a2 + b2 + c2 = 1
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
If the sum of the roots of ax2+ bx + c = 0 is equal to the sum of the ...
Let the roots be α and β
Sum of roots = -b/a
α + β = -b/a
Product of the roots = c/a
αβ = c/a
Sum of square of their reciprocal = 1/α2 + 1/β2
⇒ (α2 + β2)/(α2β2)
⇒ [(α + β)2 - 2αβ]/(α2β2)
Sum of roots = Sum of square of their reciprocal
α + β = [(α + β)2 - 2αβ]/(α2β2)
α2β2 (α + β) = (α + β)2 - 2αβ
(c/a)2 × (-b/a) = (-b/a)2 - 2c/a
-bc2/a3 = b2/a2 - 2c/a
-bc2/a3 = (b2 - 2ac)/a2
-bc2 = [a(b2 - 2ac)]
Rearranging the terms
∴ ab2 + bc2 = 2a2c
Sum of roots = -b/a
α + β = -b/a
Product of the roots = c/a
αβ = c/a
Sum of square of their reciprocal = 1/α2 + 1/β2
⇒ (α2 + β2)/(α2β2)
⇒ [(α + β)2 - 2αβ]/(α2β2)
Sum of roots = Sum of square of their reciprocal
α + β = [(α + β)2 - 2αβ]/(α2β2)
α2β2 (α + β) = (α + β)2 - 2αβ
(c/a)2 × (-b/a) = (-b/a)2 - 2c/a
-bc2/a3 = b2/a2 - 2c/a
-bc2/a3 = (b2 - 2ac)/a2
-bc2 = [a(b2 - 2ac)]
Rearranging the terms
∴ ab2 + bc2 = 2a2c
Most Upvoted Answer
If the sum of the roots of ax2+ bx + c = 0 is equal to the sum of the ...
Let the roots be α and β
Sum of roots = -b/a
α + β = -b/a
Product of the roots = c/a
αβ = c/a
Sum of square of their reciprocal = 1/α2 + 1/β2
⇒ (α2 + β2)/(α2β2)
⇒ [(α + β)2 - 2αβ]/(α2β2)
Sum of roots = Sum of square of their reciprocal
α + β = [(α + β)2 - 2αβ]/(α2β2)
α2β2 (α + β) = (α + β)2 - 2αβ
(c/a)2 × (-b/a) = (-b/a)2 - 2c/a
-bc2/a3 = b2/a2 - 2c/a
-bc2/a3 = (b2 - 2ac)/a2
-bc2 = [a(b2 - 2ac)]
Rearranging the terms
∴ ab2 + bc2 = 2a2c
Sum of roots = -b/a
α + β = -b/a
Product of the roots = c/a
αβ = c/a
Sum of square of their reciprocal = 1/α2 + 1/β2
⇒ (α2 + β2)/(α2β2)
⇒ [(α + β)2 - 2αβ]/(α2β2)
Sum of roots = Sum of square of their reciprocal
α + β = [(α + β)2 - 2αβ]/(α2β2)
α2β2 (α + β) = (α + β)2 - 2αβ
(c/a)2 × (-b/a) = (-b/a)2 - 2c/a
-bc2/a3 = b2/a2 - 2c/a
-bc2/a3 = (b2 - 2ac)/a2
-bc2 = [a(b2 - 2ac)]
Rearranging the terms
∴ ab2 + bc2 = 2a2c
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If the sum of the roots of ax2+ bx + c = 0 is equal to the sum of the squares of their reciprocals, then which one of the following relation is correct?a)ab2+ bc2= 2a2cb)ac2+ bc2= 2b2ac)ab2+ bc2= a2cd)a2+ b2+ c2= 1Correct answer is option 'A'. Can you explain this answer?
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