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How can they assume it is 4(N-L)²>0 as they didn't say the nature is real or distinct
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Thread Starter
Sugam SMHello moderators please reply it's been 2hrs
It will be responded in a reasonable time frame.
The roots of the quadratic equation will be real and distinct if the discriminant, b^2-4ac is greater than zero, the roots are equal if it is equal to zero and roots will be complex if discriminant is less than zero. The quadratic equation will have rational roots, if the value of discriminant (D) > 0 and D is not a perfect square.
Sudha Reddy
The roots of the quadratic equation will be real and distinct if the discriminant, b^2-4ac is greater than zero, the roots are equal if it is equal to zero and roots will be complex if discriminant is less than zero. The quadratic equation will have rational roots, if the value of discriminant (D) > 0 and D is not a perfect square.
Mam but we don't know soln of LMN but sum is 0 after finding b²-4ac we got 4(N-L)² but assuming D to be greater than 0 is what I am not justified. If L+M+N=0 and if N and L both are 2 the whole working will be 0 which gives D=0. Bcz in question they didn't state anything about Nature of discriminant. So if i assume L be -4 M and N be 2,2 First condition is satisfied i.e., sum is 0 but sir took Discriminant as greater than 0 which is not justified.
Thread Starter
Sugam SMThe highlighted part is where I am confused
Since it's s perfect square it is non negative. Pfa the explanation
Pavan Kumar Faculty
Since it's s perfect square it is non negative. Pfa the explanation
Yes sir both 4 and (N-L)²is perfect square so real, rational and distinct and my assumption for 2,2 is not possible. Thanks a lot sir
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