Which of the following is a quadratic equation?
\(x^2+2x+1=(4-x)^2+3\)
\(-2x^2=(5-x)\left(2x-\dfrac{2}{5}\right)\)
\((k+1)x^2+\dfrac{3}{2}x=7,\; k=-1\)
\(x^3-x^2=(x-1)^3\)
To check if an equation is quadratic, remember: a quadratic equation must have the highest power of \(x\) as 2 (like \(ax^2+bx+c=0\), where \(a \neq 0\)).
(A) Expand the right side:
\((4-x)^2 + 3 = (x^2 - 8x + 16) + 3 = x^2 - 8x + 19\).
The left side is \(x^2 + 2x + 1\). Subtract the right side from the left:
\( (x^2 + 2x + 1) - (x^2 - 8x + 19) = 10x - 18 = 0 \).
This is a linear equation (highest power of \(x\) is 1).
(B) Expand the right side:
\((5 - x)\left(2x - \dfrac{2}{5}\right) = (5)(2x) + (5)\left(-\dfrac{2}{5}\right) + (-x)(2x) + (-x)\left(-\dfrac{2}{5}\right)\).
= \(10x - 2 - 2x^2 + \dfrac{2}{5}x\).
= \(-2x^2 + \dfrac{52}{5}x - 2\).
The left side is \(-2x^2\). If we move terms, the \(-2x^2\) cancels and we are left with a linear equation.
(C) Substitute \(k = -1\):
Equation becomes \((k+1)x^2 + \dfrac{3}{2}x = 7 \Rightarrow (0)x^2 + \dfrac{3}{2}x = 7\).
So it reduces to \( \dfrac{3}{2}x = 7 \), which is a linear equation (no \(x^2\) term left).
(D) Expand the right side:
\((x-1)^3 = x^3 - 3x^2 + 3x - 1\).
The left side is \(x^3 - x^2\). Subtract RHS from LHS:
\((x^3 - x^2) - (x^3 - 3x^2 + 3x - 1) = 2x^2 - 3x + 1 = 0\).
This is a quadratic equation because the highest power of \(x\) is 2.
Therefore, the correct answer is option (D).