Solve (x-4)² + 3x² = -16x + 12: Multiple Squared Terms Equation

The formula $(a-b)^2 = a^2 - 2ab + b^2$ always gives three terms. So $(x-4)^2 = x^2 - 2(x)(4) + 4^2 = x^2 - 8x + 16$. Never forget that middle term!

Look for the pattern $a^2 + 2ab + b^2$ or $a^2 - 2ab + b^2$. In our case, $x^2 + 2x + 1 = (x+1)^2$ because the middle term is twice the product of the square roots.

Yes! When the discriminant equals zero, you get a repeated root. Here, $(x+1)^2 = 0$ means $x = -1$ is a double root - the parabola just touches the x-axis at one point.

Use the quadratic formula: $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$. For $x^2 + 2x + 1 = 0$, you get $x = \frac{-2 \pm \sqrt{4-4}}{2} = -1$.

Setting the equation to zero is the standard form for solving quadratics. It allows us to factor, complete the square, or use the quadratic formula. Always rearrange to get $ax^2 + bx + c = 0$ first!