The roots of the given equation 2(a^2 + b^2)x | vedclass

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(D) Given equation: $2(a^2 + b^2)x^2 + 2(a + b)x + 1 = 0$Comparing with $Ax^2 + Bx + C = 0$,we have:$A = 2(a^2 + b^2)$,$B = 2(a + b)$,$C = 1$The discr

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Imaginary

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(D) Given equation: $2(a^2 + b^2)x^2 + 2(a + b)x + 1 = 0$Comparing with $Ax^2 + Bx + C = 0$,we have:$A = 2(a^2 + b^2)$,$B = 2(a + b)$,$C = 1$The discriminant $D = B^2 - 4AC$ is:$D = [2(a + b)]^2 - 4 	imes 2(a^2 + b^2) 	imes 1$$D = 4(a^2 + b^2 + 2ab) - 8(a^2 + b^2)$$D = 4a^2 + 4b^2 + 8ab - 8a^2 - 8b^2$$D = -4a^2 - 4b^2 + 8ab$$D = -4(a^2 + b^2 - 2ab)$$D = -4(a - b)^2$Since $(a - b)^2 \ge 0$,it follows that $D = -4(a - b)^2 \le 0$.If $a 
eq b$,then $D < 0$,which implies the roots are imaginary.

The roots of the given equation $2(a^2 + b^2)x^2 + 2(a + b)x + 1 = 0$ are

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