Question

show that the equation 2(a2 + b2 ) x2 + 2(a+b)x + 1 =0 has no real roots , when a not equal to b .

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I HAVE ALREADY CHECKED THE EARLIER ANSWERS PROVIDED BY U BUT THERE WAS A MISTAKE IN THAT

Answer 1

Given :

The given equation is :

$2(a^2 + b^2)x^2 + 2(a+b)x + 1 = 0$

To prove:

The given equation has no real roots when a ≠ b .

Solution :

Since for a quadratic equation to have real roots , its discriminant should be positive or equal to zero .

So here :

$[2(a+b)]^2 - 4 \times 2(a^2+b^2) \times 1 \ge 0$

Or, $4(a^2+b^2+2ab) - 8a^2 - 8b^2 \ge 0$

Or , $(4a^2-8a^2) + (4b^2 -8b^2) + 8ab \ge 0$

Or, $4a^2 + 4b^ - 8ab \le 0$

Or, $a^2+ b^2 - 2ab \le 0$

Or, $(a-b)^2 \le 0$

Since the square of any number can never be less than 0 .

So the only possibility is :

$(a-b)^2 = 0$

Or, $a-b = 0$

Or, a= b

Hence , the given equation can have real roots only when a = b .

Answer 2

Step-by-step explanation:

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