Ask your own question, for FREE!
Mathematics 18 Online
OpenStudy (anonymous):

IIT mains

OpenStudy (anonymous):

Find all the integral values of a for which the quadratic equation (x-a)(x-10) +1 =0 has integral roots

OpenStudy (anonymous):

options?

OpenStudy (anonymous):

There are no options for this this came in very early in IIT . With options a new born baby would do that

OpenStudy (anonymous):

lol...then y do u ask?

OpenStudy (anonymous):

i didn't get what u mean?

OpenStudy (anonymous):

I tried to think of a approach to start this question x^2 -(a+10)x + 10a +1 =0 Since integral roots will always be rational it means that D should be a perfect square D= a^2 -20a +96 D=(a-10)^2 - 4

OpenStudy (anonymous):

Or 4 = (a-10)^2 -D So here we would need the difference between two perfect squares to be 4

OpenStudy (anonymous):

How to proceed further or is my analogy incorrect

hartnn (hartnn):

what are your A, B, C values ?

hartnn (hartnn):

i get A = 1 B = -(10+a) C = 1+10 a

OpenStudy (anonymous):

Why we need them

OpenStudy (anonymous):

m trying to approach the question in tow parts... first...i find the range of 'a' for which the equation is satisfies...that is there are real roots... and in the second part...i try to find range when the roots would be integral... after finding these range...we can take the intersection of these ranges and write the integers in the range as answer

hartnn (hartnn):

because we want \(B^2-4AC\) to be a perfect square

hartnn (hartnn):

for (x-a)(x-10) +1 =0 to have integer roots

hartnn (hartnn):

so expnd (x-a)(x-10) +1 =0

OpenStudy (anonymous):

See above i calculated it ^^

hartnn (hartnn):

a^2 -20a +96 is correct!

mathslover (mathslover):

(a-10^2 + 4 = D is also correct.

OpenStudy (anonymous):

answer is 12

mathslover (mathslover):

* (a-10)^2 + 4 = D

hartnn (hartnn):

so we need a^2 -20a +96 to have integer values

hartnn (hartnn):

thats just (a-8)(a-12) so integer values of a are just 8,12

OpenStudy (anonymous):

tanmay is half correct 12 ,8 are the answers i saw them

hartnn (hartnn):

you'll get both, 12,8

OpenStudy (anonymous):

\[a^2-20a+96=0\] completing the square...we get \[(a-10)^2=4\] a=12

hartnn (hartnn):

consider negative root too

OpenStudy (anonymous):

Yes consider negative

hartnn (hartnn):

a-10 = -2

mathslover (mathslover):

Was it so easy? :) a-10 = \(\pm \sqrt{4}\) a-10 = + 2 or a-10 =-2

hartnn (hartnn):

yep

OpenStudy (anonymous):

Yeah compared to iit level it was , they put some questions of 11th std

OpenStudy (anonymous):

In mains not in advanced

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!