The equation (1 + 2k){x^2} + (1 - 2k)x + (1 - | vedclass

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(A) The given equation is $(1 + 2k){x^2} + (1 - 2k)x + (1 - 2k) = 0$.For a quadratic equation $ax^2 + bx + c = 0$ to be a perfect square,its discrimin

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$2$

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(A) The given equation is $(1 + 2k){x^2} + (1 - 2k)x + (1 - 2k) = 0$.For a quadratic equation $ax^2 + bx + c = 0$ to be a perfect square,its discriminant $D = b^2 - 4ac$ must be equal to $0$.Here,$a = (1 + 2k)$,$b = (1 - 2k)$,and $c = (1 - 2k)$.Setting $D = 0$:$(1 - 2k)^2 - 4(1 + 2k)(1 - 2k) = 0$.Factor out $(1 - 2k)$:$(1 - 2k) [(1 - 2k) - 4(1 + 2k)] = 0$.$(1 - 2k) [1 - 2k - 4 - 8k] = 0$.$(1 - 2k) (-3 - 10k) = 0$.This gives two possible values for $k$:$1 - 2k = 0 \implies k = 1/2$.$-3 - 10k = 0 \implies k = -3/10$.Thus,there are $2$ values of $k$ for which the expression is a perfect square.

The equation $(1 + 2k){x^2} + (1 - 2k)x + (1 - 2k) = 0$ is a perfect square for how many values of $k$?

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