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If a,b are real, then the roots of the quadratic equation (a-b)x2 - 5(a+b)x -2(a-b)=0 are :-\tReal and equal.\tNon-real complex.\tReal nad unequal.\tNone of these.

Wilby Back , 10 Years ago
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Jitender Singh

Last Activity: 10 Years ago

Ans:
Hello Student,
Please find answer to your question below

a, b are real.
(a-b)x^2 - 5(a+b)x - 2(a-b) = 0
Lets find out delta.
\Delta = (-5(a+b))^{2}-4(a-b)(-2(a-b))
\Delta = 25(a+b)^{2}+8(a-b)^2
(a+b)2and (a-b)2are always positive and greater than zero.
So their sum is also positive and greater than zero.
So discriminent is positive and greater than zero.
So roots are real and unequal.

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