Learn the method, study an example, then explain your reasoning in an exam-style problem. Work on paper before opening the solution and compare each step with the marking guidance.
Factorisation and valid roots
Rearrange to zero, factorise and check every root against the original domain.
Notes
For a quadratic \(ax^2+bx+c=0\), \(a\ne0\), the zero-product rule applies after one side is zero. If \(uv=0\), then \(u=0\) or \(v=0\).
For \(a=1\), find two numbers with sum \(b\) and product \(c\). For a non-unit coefficient, split the middle term using a product \(ac\), then group terms. Check by expansion.
Never divide by an expression that could be zero without checking that case. A root of a rearranged equation is valid only if the original expression is defined.
Common mistakes
Putting each term equal to zero before obtaining a product.
Discarding zero by dividing by x.
Keeping a root that makes an original denominator zero.
Three worked examples
Example 1
Solve \(x^2-x-20=0\).
Show the reasoning
Numbers with sum \(-1\) and product \(-20\) are \(-5,4\). Hence \((x-5)(x+4)=0\), so \(x=5,-4\).
Factor out \(a\) from the first two terms before completing the square. In \(a(x-h)^2+k\), the vertex is \((h,k)\). It is a minimum if \(a>0\) and a maximum if \(a<0\).
On a restricted domain, check whether the vertex is included and evaluate the endpoints. Solving an equation requires both square-root signs.
Common mistakes
Forgetting to multiply the compensation term by a.
Reading the sign of the vertex incorrectly.
Using the unrestricted minimum on a domain that excludes the vertex.
Three worked examples
Example 1
Write \(x^2+8x+11\) in completed-square form.
Show the reasoning
\(x^2+8x+11=(x+4)^2-16+11=(x+4)^2-5\). Its minimum is \(-5\) at \(x=-4\).
Example 2
Find the maximum of \(-3x^2+12x-5\).
Show the reasoning
\(-3(x^2-4x)-5=-3(x-2)^2+7\). The square is nonnegative, so the maximum is \(7\) at \(x=2\).
Example 3
Find the range of \(2(x-1)^2+3\) for \(2\le x\le4\).
Show the reasoning
The vertex \(x=1\) is outside the domain. The function increases here; endpoint values are \(5\) and \(21\). The range is \(5\le y\le21\).
Translate root and tangency conditions into inequalities, retaining all necessary restrictions.
Notes
For real coefficients and \(a\ne0\), the discriminant is \(\Delta=b^2-4ac\). A positive value gives two distinct real roots, zero gives a repeated real root, and a negative value gives no real roots.
The roots satisfy \(\alpha+\beta=-b/a\) and \(\alpha\beta=c/a\). Two distinct positive roots require \(\Delta>0\), a positive sum and a positive product. A positive discriminant alone does not establish the signs.
To study intersections, equate the two expressions for \(y\). A tangent corresponds to a repeated intersection root. If a parameter can make \(a=0\), handle that linear case separately.
Common mistakes
Using a non-strict inequality when roots must be distinct.
Treating two real roots as necessarily positive.
Ignoring a parameter value that turns a quadratic into a linear equation.
Three worked examples
Example 1
How many real roots does \(2x^2+4x+5=0\) have?
Show the reasoning
\(\Delta=4^2-4(2)(5)=-24<0\), so there are no real roots.
Example 2
Find \(k\) if \(x^2-6x+k=0\) has a repeated root.
Show the reasoning
\(36-4k=0\) gives \(k=9\). The equation is \((x-3)^2=0\), with repeated root \(3\).
Example 3
For which \(r\) does \(x^2-4x+r=0\) have two distinct positive roots?
Show the reasoning
The discriminant gives \(r<4\). The sum is \(4>0\); the product requires \(r>0\). Together, \(0
Use roots and signs to find intervals, then impose additional constraints.
Notes
First rearrange the inequality so that one side is zero. Locate its real roots and determine the sign on each interval. For a quadratic with positive leading coefficient and two distinct roots, it is negative between the roots and positive outside.
Include a root for \(\le\) or \(\ge\), and exclude it for \(<\) or \(>\). With a negative leading coefficient, reverse the sign pattern. A repeated root does not change the sign.
Intersect the solution set with every domain restriction. If the variable counts objects, list only integers in the resulting interval. Multiplying by an unknown-sign expression can reverse an inequality; split cases instead.
Common mistakes
Reporting only the roots instead of an interval.
Using closed endpoints for a strict inequality.
Forgetting the integer or domain restriction.
Three worked examples
Example 1
Solve \(x^2-x-6<0\).
Show the reasoning
\((x-3)(x+2)<0\). The upward-opening quadratic is negative between its roots: \(-2
Example 2
Solve \(-x^2+4x+5\ge0\).
Show the reasoning
Equivalently \((x-5)(x+1)\le0\), so \(-1\le x\le5\). Multiplication by \(-1\) reverses the inequality.
Example 3
An integer \(n\ge0\) satisfies \((n-2)(n-6)\le0\). List its values.
Show the reasoning
The real interval is \(2\le n\le6\). The allowable integers are \(2,3,4,5,6\), including both endpoints.
Substitute, solve and reconstruct all coordinate pairs; recognise a tangent.
Notes
A linear equation lets you express one variable in terms of the other. Substitute into the quadratic, simplify to zero and solve. Then use the linear equation to recover the matching second coordinate for each root.
Check each pair in both original equations. Do not combine the first coordinate of one solution with the second coordinate of another.
The resulting quadratic describes the intersections. Two distinct roots give two points, a repeated root gives one tangent point, and no real roots mean no real intersection.
Common mistakes
Stopping after finding x.
Mixing coordinates from different solutions.
Claiming a tangent without showing a repeated intersection root.
Three worked examples
Example 1
Solve \(y=x+1\) and \(y=x^2-5\).
Show the reasoning
\(x^2-x-6=(x-3)(x+2)=0\). Thus the points are \((3,4)\) and \((-2,-1)\).
Example 2
Find the intersections of \(y=2x\) and \(x^2+y^2=45\).
Show the reasoning
Substitution gives \(5x^2=45\), so \(x=\pm3\). The matched pairs are \((3,6)\), \((-3,-6)\).
Example 3
Show that \(y=4x-4\) is tangent to \(y=x^2\).
Show the reasoning
\(x^2-4x+4=(x-2)^2=0\). There is one repeated intersection at \((2,4)\), so the line is tangent.