Quadratic Functions and Equations in One Variable · Form 4
Quadratic Functions and Equations in One Variable: Common Mistakes
The mistakes that quietly cost marks in Quadratic Functions and Equations in One Variable, and how to avoid each one in the SPM exam.
In our experience teaching Quadratic Functions and Equations in One Variable, most lost marks come from a small set of repeated slips, not from a lack of understanding. Here they are, with the fix for each.
Mistakes to avoid
- Forgetting the ± when taking a square root, so one root goes missing
- Sign errors when the coefficient a is negative
- Reading the discriminant’s sign wrong and miscounting the roots
Six more slips that quietly cost marks
- What students write: solving x2 + 2x = 15 as x(x + 2) = 15, then x = 15 or x + 2 = 15. -> Why it loses marks: factorisation only works when one side is 0; a product equal to 15 says nothing about each factor. -> Correct working: rearrange to x2 + 2x - 15 = 0, factorise (x + 5)(x - 3) = 0, so x = -5 or x = 3.
- What students write: from x2 = 3x, dividing both sides by x to get x = 3. -> Why it loses marks: dividing by x throws away the root x = 0. -> Correct working: x2 - 3x = 0, factorise x(x - 3) = 0, so x = 0 or x = 3.
- What students write: reading (x + 4)2 - 1 as a turning point at x = 4. -> Why it loses marks: the form is a(x - h)2 + k, and x + 4 means h = -4, not +4. -> Correct working: (x + 4)2 - 1 has turning point (-4, -1).
- What students write: completing the square on 2x2 - 8x + 5 as (x - 2)2 + ... without touching the 2. -> Why it loses marks: you must factor a out of the x-terms first, or the coefficient of x2 is wrong. -> Correct working: 2(x2 - 4x) + 5 = 2(x - 2)2 - 8 + 5 = 2(x - 2)2 - 3.
- What students write: giving the axis of symmetry as x = b/2a. -> Why it loses marks: the correct formula has a negative sign, so the sign of the answer flips. -> Correct working: axis of symmetry is x = -b/2a; for y = x2 - 6x + 5 it is x = -(-6)/2 = 3.
- What students write: solving a quadratic and stopping after the first factor gives one value. -> Why it loses marks: a positive discriminant means two roots, and the second one carries its own mark. -> Correct working: from (x - 2)(x - 3) = 0 write both x = 2 and x = 3.
A ten-second self-check before you move on
After solving, substitute each root back into the original equation to confirm it gives 0, and check the number of roots against the sign of b2 - 4ac. For a turning point, expand your completed-square form once to confirm it returns the original quadratic.
These quick checks catch dropped roots, sign flips and completing-square errors before they cost you marks.
Source:DSKP KSSM Mathematics Form 4 and 5 (Versi English)
Frequently asked questions
I keep getting the turning point's x-coordinate with the wrong sign. How do I check it fast?
Remember the form is a(x - h)2 + k, so h is whatever makes the bracket zero. For (x + 4)2, set x + 4 = 0, giving x = -4.
Substituting x = -4 back should make the squared term vanish; if it does not, your sign is wrong. This one-second check catches almost every sign slip.
Is it ever okay to divide both sides of a quadratic by x to simplify?
No. Dividing by x silently assumes x is not zero and deletes the root x = 0, which is often one of the two marks.
Instead move everything to one side and factor out x. For x2 = 3x, write x2 - 3x = 0 then x(x - 3) = 0, keeping both x = 0 and x = 3.
How do I avoid rounding errors when I use the quadratic formula?
Keep the surd in exact form for as long as you can and only round at the very last step. For x = [4 +/- sqrt(12)] / 2, do not round sqrt(12) to 3.5 early; carry sqrt(12) = 3.4641 and finish as x = 3.73 or 0.27.
Rounding mid-calculation shifts the final answer past the accepted two decimal places.