For how many values of k is the equation (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 discriminan

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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 $D = 0$,we get $(1 - 2k)^2 - 4(1 + 2k)(1 - 2k) = 0$.Factoring out $(1 - 2k)$,we have $(1 - 2k) \cdot [(1 - 2k) - 4(1 + 2k)] = 0$.$(1 - 2k) \cdot [1 - 2k - 4 - 8k] = 0$.$(1 - 2k) \cdot (-3 - 10k) = 0$.This gives two possible values for $k$: $1 - 2k = 0 \implies k = \frac{1}{2}$ and $-3 - 10k = 0 \implies k = -\frac{3}{10}$.However,if $k = -\frac{1}{2}$,the coefficient of $x^2$ becomes $0$,and the equation is no longer quadratic. Checking $k = \frac{1}{2}$,the equation becomes $2x^2 = 0$,which is a perfect square. Checking $k = -\frac{3}{10}$,the equation becomes $(1 - 0.6)x^2 + (1 + 0.6)x + (1 + 0.6) = 0$,i.e.,$0.4x^2 + 1.6x + 1.6 = 0$,which simplifies to $x^2 + 4x + 4 = (x+2)^2 = 0$,which is also a perfect square.Thus,there are $2$ such values of $k$.

For how many values of $k$ is the equation $(1 + 2k)x^2 + (1 - 2k)x + (1 - 2k) = 0$ a perfect square?

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