Theory of Equations 5 Question 5

5. Let a,b,c be real numbers, a0. If α is a root of a2x2+bx+c=0,β is the root of a2x2bxc=0 and 0<α<β, then the equation a2x2+2bx+2c=0 has a root γ that always satisfies

(1989,2M)

(a) γ=α+β2

(b) γ=α+β2

(c) γ=α

(d) α<γ<β

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Answer:

Correct Answer: 5. (d)

Solution:

  1. Since, α is a root of a2x2+bx+c=0

a2α2+bα+c=0

and β is a root of a2x2bxc=0

a2β2bβc=0

Let f(x)=a2x2+2bx+2c

f(α)=a2α2+2bα+2c

=a2α22a2α2=a2α2

[from Eq. (i)]

and f(β)=α2β2+2bβ+2c

=a2β2+2a2β2=3a2β2 [from Eq. (ii)] 

f(α)f(β)<0

f(x) must have a root lying in the open interval (α,β).

α<γ<β



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