Second Degree Equations: Roots and Discriminant
What the exam tests
Second Degree Equations covers solving quadratics by factorisation, completing the square and the quadratic formula, interpreting the discriminant, and forming an equation from given roots. Word problems often require rejecting a root that does not fit the context.
Core rules
- Standard form: ax²+bx+c=0, with a≠0.
- Quadratic formula: x = [−b ± √(b²−4ac)]/(2a).
- Discriminant D=b²−4ac: D>0 gives two distinct real roots; D=0 gives equal roots; D<0 gives no real roots.
- Sum of roots = −b/a; product of roots = c/a.
- If the roots are α and β, the equation is x²−(α+β)x+αβ=0.
- Completing the square rewrites ax²+bx+c as a perfect-square part plus a constant.
Worked example
Solve 2x²−7x+3=0. Here D=(−7)²−4(2)(3)=49−24=25. Thus x=(7±5)/4, giving x=12/4=3 or x=2/4=1/2. Check: sum of roots =3+1/2=7/2=−b/a, and product =3/2=c/a.
Traps
A sign error in −b or in b²−4ac changes both roots. Do not divide by x and lose the root x=0. D<0 means no real roots in this syllabus, not two negative roots. When forming an equation from roots, use x²−(sum)x+product, not x²+(sum)x+product. In word problems, reject negative lengths or times even if they satisfy the algebra.