If the equation x 2 - 2px+q = 0 has two equal roots, then show that the eqn (1+y) x 2 -2(p+y)x + (q+y) = 0 will have it's root real and distinct only when y is -ve and p is not unity.

Question

Pratham Ashish · Answer

x^2-2px+q=0 since this eqn hav equal roots, b^2 = 4ac,in this case using this condition we can write 4*p^2 = 4*q => p^2 = q...........(1) now for 2nd eqn (1+y)x2 -2(p+y)x + (q+y) = 0, we hav to make roots real and dinstinct so condition wud be (b^2 - 4ac) > 0, then eqn wud be (4(p+y))^2 - 4(q+y)(1+y) >0 when u solve this eqn anp put p^2 in place of q u will get a final eqn that is y*(2p - p^2 -1)>0 now u can see at p=1 , the left hand side will become 0 so p can't be unity, and the quadratic in p that is (2p - p^2 -1) is always -ve for all values of p as it is a parabola with concavity downward touching x axis at only one point that is 1 so for making the final expression of y*(2p - p^2 -1) positive , y should be -ve so that product become > 0...... Please feel free to post as many doubts on our disucssion forum as you can. If you find any question difficult to understand - post it here and we will get you the answer and detailed solution very quickly. We are all IITians and here to help you in your IIT JEE preparation. All the best . Regards, Askiitians Experts Pratham IIT - Delhi

Thank you for registering.

Register Now and Win Upto 25% Scholorship for a Full Academic Year !

Enter Your Details

Registration done!

Guest

If the equation x 2 - 2px+q = 0 has two equal roots, then show that the eqn (1+y) x 2 -2(p+y)x + (q+y) = 0 will have it's root real and distinct only when y is -ve and p is not unity.

If the equation x²- 2px+q = 0 has two equal roots, then show that the eqn (1+y) x² -2(p+y)x + (q+y) = 0 will have it's root real and distinct only when y is -ve and p is not unity.

1 Answers

Think You Can Provide A Better Answer ?

Other Related Questions on Algebra

ASK QUESTION

Answer and earn gift vouchers

View courses by askIITians

Design classes One-on-One in your own way with Top IITians/Medical Professionals

Complete Self Study Package designed by Industry Leading Experts

Live 1-1 coding classes to unleash the Creator in your Child

a Complete All-in-One Study package Fully Loaded inside a Tablet!

Register Now & Win Upto 25% Scholorship

Thank you for registering.

Register Now and Win Upto 25% Scholorship for a Full Academic Year !

Enter Your Details

Registration done!

Guest

IIT JEE Courses

One Year IIT Programme

Two Year IIT Programme

Crash Course

NEET Courses

One Year NEET Programme

Two Year NEET Programme

One to One

School Board

Live Online classes

Class 11 & 12

Class 9 & 10

Class 6, 7 & 8

Test Series

JEE test series

NEET test series

C.B.S.E test series

Complete Self Study Packages

FULL COURSE

For class 12th

For class 11th

If the equation x 2 - 2px+q = 0 has two equal roots, then show that the eqn (1+y) x 2 -2(p+y)x + (q+y) = 0 will have it's root real and distinct only when y is -ve and p is not unity.

If the equation x2 - 2px+q = 0 has two equal roots, then show that the eqn (1+y) x2 -2(p+y)x + (q+y) = 0 will have it's root real and distinct only when y is -ve and p is not unity.

1 Answers

Think You Can Provide A Better Answer ?

Other Related Questions on Algebra

ASK QUESTION

Answer and earn gift vouchers

View courses by askIITians

Design classes One-on-One in your own way with Top IITians/Medical Professionals

Complete Self Study Package designed by Industry Leading Experts

Live 1-1 coding classes to unleash the Creator in your Child

a Complete All-in-One Study package Fully Loaded inside a Tablet!

Register Now & Win Upto 25% Scholorship

If the equation x²- 2px+q = 0 has two equal roots, then show that the eqn (1+y) x² -2(p+y)x + (q+y) = 0 will have it's root real and distinct only when y is -ve and p is not unity.