If (1/ p ), (1/ q ), (1/ r ) are in A.P. and p and r be different having same sign, then the roots of the equation px 2+2 qx + r =0 will be

Question

If (1/ p ), (1/ q ), (1/ r ) are in A.P. and p and r be different having same sign, then the roots of the equation px 2+2 qx + r =0 will be (A) real (B) equal (C) imaginary (D) real and distinct

Tardigrade Admin · Accepted Answer

Since, (1/p), (1/q), (1/r) in A P ⇒ q=(2 p r/p+r) Now, D=4 q2-4 p r =-4[p r-((2 p r/p+r))2] =-(p r)[2((p-r/p+r))]2 [From equation (i)] Since, pr >0, p ≠ r given, D ≠ 0 and D <0. Hence, the roots are imaginary.

Q. If p1​,q1​,r1​ are in A.P. and p and r be different having same sign, then the roots of the equation px2+2qx+r=0 will be

real

equal

imaginary

real and distinct

Since, p1​,q1​,r1​ in AP⇒q=p+r2pr​ Now, D=4q2−4pr=−4[pr−(p+r2pr​)2] =−(pr)[2(p+rp−r​)]2 [From equation (i)] Since, pr>0,p=r given, D=0 and D<0. Hence, the roots are imaginary.

Q. If $\frac{1}{p}, \frac{1}{q}, \frac{1}{r}$ are in A.P. and $p$ and $r$ be different having same sign, then the roots of the equation $p x^{2} + 2 q x + r = 0$ will be