# Algebraic manipulation of a quadratic equation to an alternative form

I have a quadratic equation that I have attempted to manipulate algebraically but still don't have a reasonable solution.

Given a quadratic equation as follows

$$a(x-k)^2 + b(k-c)^2$$

how could this be expressed in the form

$$d(e-k)^2 + f$$ such that the terms $$d$$, $$e$$ and $$f$$ do not involve $$k$$, preferably using completing the square approach.

Given $$a(x-k)^2 + b(k-c)^2 = d(e-k)^2+f$$, you're right, the easiest way is to complete the square.

First, we need to expand the expression:

$$a(x - k)^2 + b(k - c)^2 = ax^2 - 2axk + ak^2 + bk^2 - 2bck + bc^2$$

Then, we group within different powers of $$k$$:

$$= ak^2 + bk^2 - 2axk - 2bck + ax^2 + bc^2$$ $$= (a+b)k^2 - 2(ax + bc)k + ax^2 + bc^2$$

Next, we factor out $$(a+b)$$ from the first two terms and complete the square using just those two terms:

$$= (a+b)\left[k^2 - 2\left(\frac{ax + bc}{a+b}\right)k\right] + ax^2 + bc^2$$

$$= (a+b)\left[k^2 - 2\left(\frac{ax + bc}{a+b}\right)k + \left(\frac{ax + bc}{a+b}\right)^2 \right] + ax^2 + bc^2 - \frac{(ax + bc)^2}{a+b}$$

$$= (a+b)\left(k - \frac{ax + bc}{a+b}\right)^2 + ax^2 + bc^2 - \frac{(ax + bc)^2}{a+b}$$

At this point, we can reverse the order of the expression within the square (to look more like the expression we're trying to replicate):

$$= (a+b)\left(\frac{ax + bc}{a+b} - k\right)^2 + ax^2 + bc^2 - \frac{(ax + bc)^2}{a+b}$$

With a bit more manipulation, we can further simplify the expression after the square:

$$= (a+b)\left(\frac{ax + bc}{a+b} - k\right)^2 + \frac{(a+b)(ax^2 + bc^2) - (ax + bc)^2}{a+b}$$

$$= (a+b)\left(\frac{ax + bc}{a+b} - k\right)^2 + \frac{a^2x^2 + abx^2 + abc^2 + b^2c^2 - a^2x^2 - 2abcx + b^2c^2}{a+b}$$

$$= (a+b)\left(\frac{ax + bc}{a+b} - k\right)^2 + \frac{abx^2 + abc^2 - 2abcx}{a+b}$$

$$= (a+b)\left(\frac{ax + bc}{a+b} - k\right)^2 + \frac{ab}{a+b}(x^2 - 2abcx + c^2)$$

$$= (a+b)\left(\frac{ax + bc}{a+b} - k\right)^2 + \frac{ab}{a+b}(x - c)^2$$

Comparing against the form we want: $$d(e-k)^2+f$$, we have:

$$d = a + b ,$$ $$e = \frac{ax + bc}{a+b} ,$$ $$f = \frac{ab}{a+b}(x - c)^2 .$$

• Absolutely! I was missing the critical step where $(a+b)$ is factored out so that it would be a quadratic on $k$. Jan 21, 2021 at 8:32
• You can leave it in and still complete the square, it's just a lot easier to factor it out! Jan 21, 2021 at 18:01