Try each one on paper first, then check a single step. That is worth far more than reading a finished solution.
Example 1 — completing the square when the leading coefficient is not 1
Express y=2x2+8x+5 in the form a(x+p)2+q, and hence state its minimum value and the value of x at which it occurs.
- 1y=2(x2+4x)+5Take the 2 out of the first two terms only. The constant stays outside.
- 2x2+4x=(x+2)2−4Half of 4 is 2, and (x+2)2 carries an extra +4 that must be removed.
- 3y=2[(x+2)2−4]+5Put it back inside the bracket.
- 4y=2(x+2)2−8+5=2(x+2)2−3Multiply the −4 by the 2 on the way out. This is the step people miss.
- 5Minimum value −3, at x=−2.(x+2)2 is smallest when it is zero, which is at x=−2; the value there is q=−3.
Example 2 — using the discriminant to find a parameter
Find the values of k for which x2+kx+9=0 has equal roots.
- 1Equal roots means b2−4ac=0.A repeated root is exactly the zero-discriminant case.
- 2Here a=1, b=k, c=9, so k2−36=04ac=4×1×9=36.
- 3k2=36⇒k=±6Take the square root of both sides — and keep both signs.
- 4So k=6 or k=−6.Check: x2+6x+9=(x+3)2, a repeated root at x=−3.
Example 3 — working with the roots without finding them
The roots of 2x2−5x+1=0 are α and β. Find α2+β2.
- 1α+β=−ab=−2−5=25b is −5, so the two minus signs cancel.
- 2αβ=ac=21The product is simply c over a.
- 3α2+β2=(α+β)2−2αβThe identity to memorise. Expanding (α+β)2 gives an extra 2αβ that must be removed.
- 4=(25)2−2(21)=425−1=421The roots themselves are irrational, which is why this route is quicker.