Falling Behind in Studies? Join Our Performance Improvement Batch. Enroll For Free
One of our academic counsellors will contact you within 1 working day.
Click to Chat
1800-1023-196
+91-120-4616500
CART 0
Use Coupon: CART20 and get 20% off on all online Study Material
Welcome User
OR
LOGIN
Complete Your Registration (Step 2 of 2 )
Live 1-1 coding classes to unleash the creator in your Child
Q1. The roots alpha and betaof quadratioc equation pxsquare+qx+r are real and of opposite sign . the roots of alpha(x-beta)square+beta(x-alpha)square=0are a) positive b) neagative c) real and of opposite sign d) imaginary
α(x-β)2+β(x-α)2=0 ---(1) Given α and β are real and of opposite sign Therefore αβ= -ve From equation 1 α(x2+β2-2xβ) + β(x2+α2-2αx)=0 (α+β)x2-4αβx+αβ(α+β)=0 product of the roots = αβ(α+β)/(α+β)= αβ = -ve discriminant Δ=16 α2β2-4 αβ( α+β)2 as 16 α2β2 and ( α+β)2 are positive and αβ is negative, Δ is positive so the roots of the equation are real and opposite in sign. Ans: c
α(x-β)2+β(x-α)2=0 ---(1)
Given α and β are real and of opposite sign
Therefore αβ= -ve
From equation 1
α(x2+β2-2xβ) + β(x2+α2-2αx)=0
(α+β)x2-4αβx+αβ(α+β)=0
product of the roots = αβ(α+β)/(α+β)= αβ = -ve
discriminant Δ=16 α2β2-4 αβ( α+β)2
as 16 α2β2 and ( α+β)2 are positive and αβ is negative, Δ is positive
so the roots of the equation are real and opposite in sign.
Ans: c
Post Question
Dear , Preparing for entrance exams? Register yourself for the free demo class from askiitians.
points won -