IIT mains
Find all the integral values of a for which the quadratic equation (x-a)(x-10) +1 =0 has integral roots
options?
There are no options for this this came in very early in IIT . With options a new born baby would do that
lol...then y do u ask?
i didn't get what u mean?
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
Or 4 = (a-10)^2 -D So here we would need the difference between two perfect squares to be 4
How to proceed further or is my analogy incorrect
what are your A, B, C values ?
i get A = 1 B = -(10+a) C = 1+10 a
Why we need them
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
because we want \(B^2-4AC\) to be a perfect square
for (x-a)(x-10) +1 =0 to have integer roots
so expnd (x-a)(x-10) +1 =0
See above i calculated it ^^
a^2 -20a +96 is correct!
(a-10^2 + 4 = D is also correct.
answer is 12
* (a-10)^2 + 4 = D
so we need a^2 -20a +96 to have integer values
thats just (a-8)(a-12) so integer values of a are just 8,12
tanmay is half correct 12 ,8 are the answers i saw them
you'll get both, 12,8
\[a^2-20a+96=0\] completing the square...we get \[(a-10)^2=4\] a=12
consider negative root too
Yes consider negative
a-10 = -2
Was it so easy? :) a-10 = \(\pm \sqrt{4}\) a-10 = + 2 or a-10 =-2
yep
Yeah compared to iit level it was , they put some questions of 11th std
In mains not in advanced
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