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Question

If a and b are real and a b then show that the roots of the equation (ab)x2+5(a+b)x2(ab)=0  are real and unequal. 


Solution

The given equation is (ab)x2+5(a+b)x2(ab)=0

Given, a, b are real and ab.

Then, Discriminant (D) =b24ac

=[5(a+b)]24(ab)(2(ab))

=25(a+b)2+(ab)2

We know that the square of any integer is always positive that is, greater than zero.

Hence, (D) =b24ac0

As given, a, b are real and ab.

Therefore,

= 25(a+b)2+(ab)2>0=D>0

Therefore, the roots of this equation are real and unequal.
 


Mathematics
Secondary School Mathematics X
Standard X

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