1.1 Quadratics

Rewrite ax²+bx+c as a(x−h)²+k by factoring a and adding and subtracting the required square. The vertex is (h,k), and the sign of a tells whether it is a minimum or maximum.

Keep the factor outside the square: ax²+bx+c=a[x²+(b/a)x]+c. Completing the square is also the quickest route to a quadratic equation’s exact roots and range.

x²−6x+5=(x−3)²−4, so the graph has minimum −4 at x=3 and range y≥−4.

The constant changes when the square is completed; do not write (x−3)²+5, which expands to a different quadratic.

For ax²+bx+c=0, the discriminant Δ=b²−4ac. Δ>0 gives two distinct real roots, Δ=0 one repeated real root, and Δ<0 no real roots.

Use the discriminant when the question asks how many roots, a tangent condition or a parameter range. It is often cleaner than applying the quadratic formula twice.

For x²−4x+k=0, Δ=16−4k. Two real roots require k<4; tangency occurs at k=4.

Δ>0 does not mean both roots are positive; root signs require additional information such as sum and product.

A quadratic equation can be solved by factorising, completing the square or using x=(−b±√(b²−4ac))/(2a). All methods should return the same roots.

Factorise when integer or simple factors are visible; use the formula for a general or parameterised quadratic; complete the square when the vertex or inequality range is needed.

For 2x²+3x−2=0, factorising gives (2x−1)(x+2)=0, so x=1/2 or −2. The formula is a reliable check.

After multiplying or dividing an equation, preserve equivalence; taking a square root too early can lose the ± branch.

For two equations, eliminate one variable by substitution or a linear combination, solve the resulting equation, then back-substitute and verify in both original equations.

If one equation is quadratic, each valid root may produce a different pair. Keep every algebraic branch until the original equations or stated domain reject it.

From y=2x+1 and x²+y²=25, substitute to obtain x²+(2x+1)²=25; solve for both possible x values, then calculate y for each.

Solving only the positive square-root branch can omit a valid intersection, while numerical rounding can make a true pair appear not to satisfy the equations.

If an equation contains x² and x⁴, set u=x² and solve the resulting quadratic in u. Only values u≥0 correspond to real x, and each positive u gives x=±√u.

Make the substitution explicit, solve in the new variable, apply its domain restriction, then reverse the substitution and check the original equation.

x⁴−5x²+4=0 becomes u²−5u+4=0, so u=1 or 4 and x=±1,±2.

A negative solution for u cannot produce a real x, and replacing x² by u does not mean x itself equals u.