Skip to content
dronacharya.ai
← JEE Main 2014 · 6 Apr · Shift 1
MathematicsSingle correct+4 / -1Question 64 of 90

JEE Main 2014 · 6 Apr · Shift 1, Mathematics question 64

Let α\alpha and β\beta be the roots of equation px2+qx+r=0px^2+qx+r=0, p≠0p\ne0. If pp, qq, rr are in A.P. and 1α+1β=4\dfrac1\alpha+\dfrac1\beta=4, then the value of ∣α−β∣|\alpha-\beta| is :

Choose one option
Answer and worked solution

Official answer: B

Relations. 1α+1β=−qr=4\frac1\alpha+\frac1\beta=-\frac qr=4 gives q=−4rq=-4r. The AP gives p=2q−r=−9rp=2q-r=-9r.

Roots. α+β=−qp=−49\alpha+\beta=-\frac qp=-\frac49 and αβ=rp=−19\alpha\beta=\frac rp=-\frac19. So
(α−β)2=1681+49=5281,∣α−β∣=2139.(\alpha-\beta)^2=\frac{16}{81}+\frac49=\frac{52}{81},\qquad|\alpha-\beta|=\frac{2\sqrt{13}}9.

Topics:Roots and coefficientsArithmetic progression
← Previous questionNext question →