Quadratic Functions & Discriminant

Secondary Additional Mathematics · Study guide

Connect the turning point, real roots and sign of a quadratic. Learn when to complete the square, when to use the discriminant, and how to state a complete solution.

Concept → worked examples → practice10 min read · practice linked below

One quadratic, three useful forms

A quadratic question may ask for a minimum value, the number of real roots, or a range of values satisfying an inequality. These are connected questions about the same parabola. The useful first move is to choose a form that reveals the information you need.

Choose the form to match the question
Form	What it reveals
$ax^2+bx+c$ , where $a\ne0$	The coefficients for $D=b^2-4ac$ , and the y-intercept $(0,c)$ .
$a(x-h)^2+k$	The turning point $(h,k)$ and axis of symmetry $x=h$ .
$a(x-r)(x-s)$ , when real roots exist	The roots $r,s$ , useful for a sign diagram.

Before you start: be comfortable expanding brackets, factorising quadratics and solving linear inequalities. A root is an x-value for which the expression equals zero. A turning point is a coordinate pair, while a maximum or minimum value is a y-value.

Completing the square: find the turning point

Use $x^2+px=(x+\frac p2)^2-(\frac p2)^2$ . If the coefficient of $x^2$ is not 1, factor it out of the x-terms first. Keep the subtraction inside the bracket until you multiply back.

In $y=a(x-h)^2+k$ , the square is always at least zero. Therefore, if $a>0$ , the minimum value is $k$ . If $a<0$ , the maximum value is $k$ . Both occur at $x=h$ .

Example 1: a minimum with a non-unit coefficient

Express $3x^2-18x+22$ in completed-square form. State its minimum value over all real $x$ and where it occurs.

Factor 3 out of the first two terms: $3(x^2-6x)+22.$
Half of $-6$ is $-3$ , and $(x-3)^2=x^2-6x+9$ . Subtract $9$ inside the bracket to keep the expression unchanged: $3\big[(x-3)^2-9\big]+22.$
Multiply the whole bracket by 3 and simplify: $3(x-3)^2-27+22=3(x-3)^2-5.$

Since $3(x-3)^2\ge0$ , the expression is at least $-5$ , with equality when $x=3$ . The minimum is $-5$ at $x=3$ , and the turning point is $(3,-5)$ .

Check by expanding: $3(x^2-6x+9)-5=3x^2-18x+22$ .

The dashed line passes through the turning point. The graph’s lowest y-value is −5, reached at x = 3.

In a modelling question, check the permitted x-values. If $x=3$ is outside the stated interval, this turning point cannot give the required minimum. On a closed interval, compare endpoint values. An excluded endpoint may give only a limiting value, so a minimum need not be attained.

The discriminant: count real roots

For $ax^2+bx+c=0$ , where $a\ne0$ , the quadratic formula is

x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}.

The expression under the square root is the discriminant, written $D=b^2-4ac$ . Its sign determines how many real roots the equation has.

Discriminant and the x-axis
Condition	Real roots	Graph
$D>0$	Two distinct real roots	Crosses the x-axis twice
$D=0$	Two equal roots, one distinct x-value	Touches the x-axis at its turning point
$D<0$	No real roots	Does not meet the x-axis

“Has real roots” allows equality, so use $D\ge0$ . “Two distinct real roots” requires $D>0$ . There is no need to calculate the roots when the question asks only how many there are.

Positive for every real x: require $a>0$ and $D<0$ . The curve opens upwards and never meets the axis.

Negative for every real x: require $a<0$ and $D<0$ . For “non-negative” or “non-positive”, allow $D=0$ as well.

Example 2: find a parameter for strict positivity

Find the values of $k$ for which $2x^2+4x+k>0$ for every real x.

The leading coefficient is $2>0$ , so the parabola opens upwards. To be strictly above the x-axis for every real input, it must neither cross nor touch that axis. Therefore it must have no real roots, which requires $D<0$ :

\begin{aligned}D&=4^2-4(2)(k)\\&=16-8k.\end{aligned}

Add $8k$ to both sides, then divide by the positive number $8$ ; the inequality direction stays the same:

\begin{aligned}16-8k&<0\\16&<8k\\2&<k.\end{aligned}

Thus $k>2$ . The boundary $k=2$ is excluded because then $2x^2+4x+2=2(x+1)^2$ , which equals zero at $x=-1$ .

Completing the square confirms this: $2(x+1)^2+k-2$ has minimum $k-2$ , so strict positivity requires $k-2>0$ .

Explore: move the curve through the x-axis

For $y=x^2-4x+c=(x-2)^2+c-4$ , changing $c$ moves the curve vertically. Predict what happens when $c$ reaches 4, then test it.

Constant c:3

Use the slider or arrow keys. Values run from 0 to 8 in steps of 0.5.

The axis of symmetry stays at x = 2. Only the height changes.

c = 3. Turning point (2, −1). Minimum −1. Discriminant D = 4. Two distinct real roots: x = 1 and x = 3.

The two methods agree because the turning-point y-coordinate is $-D/(4a)$ . For this family, $a=1$ , so raising the minimum through zero changes $D$ from positive to zero to negative.

Quadratic inequalities: find intervals, not just roots

Rearrange so that zero is on one side. Find the boundary roots, then determine the sign on each interval using a sketch or test values. Include a boundary only when equality is allowed.

Example 3: a non-positive quadratic

Solve $x^2-3x\le10$ and show the solution on a number line.

Subtract $10$ from both sides, then factorise. We need two numbers with product $-10$ and sum $-3$ , namely $2$ and $-5$ :

\begin{aligned}x^2-3x-10&\le0\\ (x+2)(x-5)&\le0.\end{aligned}

The product is zero at $x=-2$ and $x=5$ . Within each interval between these roots, neither factor changes sign. We can therefore determine the sign of the product by testing one point in each interval:

At $x=-3$ , both factors are negative, so their product is positive.
At $x=0$ , the factors have opposite signs, so the product is negative.
At $x=6$ , both factors are positive, so their product is positive.

We need a product less than or equal to zero. It is negative between the roots, and equality includes both endpoints. On the number line, shade the segment from $-2$ to $5$ , using filled circles at both ends. Therefore $-2\le x\le5$ .

Closed circles include equality. With a strict inequality, use open circles and exclude both endpoints.

A repeated root needs care: $(x-2)^2\ge0$ for every real x, but $(x-2)^2\le0$ only when $x=2$ . The sign does not change across a repeated root.

Line–curve intersections: form a new quadratic

At an intersection, the line and curve have the same x- and y-values. Equate their expressions for y. Apply the discriminant to the resulting equation in x. For a line $y=mx+n$ and a parabola $y=ax^2+bx+c$ , it is

ax^2+(b-m)x+(c-n)=0.

A positive discriminant gives two intersection points. Zero gives a tangent, with one repeated x-value. A negative discriminant gives no intersection. If the question says “intersects” without requiring two distinct points, include the tangent case.

Example 4: find the tangent in a family of lines

Consider the family of lines $y=2x+k$ and the curve $y=x^2-4x+7$ . Find $k$ when the line is tangent and find the point of contact.

Equate y-values and rearrange:

\begin{aligned}x^2-4x+7&=2x+k\\x^2-6x+7-k&=0.\end{aligned}

The resulting quadratic gives the x-coordinates of the intersections. A tangent has one point of contact, so this quadratic must have a repeated root. Its discriminant is therefore zero:

\begin{aligned}(-6)^2-4(1)(7-k)&=0\\8+4k&=0\\k&=-2.\end{aligned}

Substitute $k=-2$ : $x^2-6x+9=(x-3)^2=0$ , giving $x=3$ . On the line, $y=2(3)-2=4$ . The tangent is $y=2x-2$ , touching at $(3,4)$ .

Check the contact point in both original equations: the line gives $2(3)-2=4$ , and the curve gives $3^2-4(3)+7=4$ . The intersection equation has the repeated root $x=3$ , confirming tangency. The same discriminant shows two intersections for $k>-2$ , and none for $k<-2$ . Using the curve’s original discriminant would only tell us about its intersections with the x-axis.

Common errors and quick checks

Forgetting the outside multiplier. In $3[(x-3)^2-9]$ , the correction is $-27$ . Expand your completed-square form to check it.
Misreading a shifted square. $(x+3)^2$ becomes zero at $x=-3$ , so its axis is $x=-3$ .
Dropping a sign in the discriminant. Substitute with brackets, especially when $b$ or $c$ is negative.
Using $D<0$ alone for “always positive”. Check $a>0$ too. A downward-opening curve with no roots is always negative.
Writing only the boundary roots. An inequality usually needs intervals. Test a point and check whether equality is included.
Assuming a parameter leaves a quadratic. If the coefficient of $x^2$ can become zero, handle that value separately before applying the discriminant.

Practise with our topic questions

Apply the methods in this guide using our A-Maths quadratic functions and inequalities practice questions. The topic bank brings the questions and their existing worked solutions together, so you can attempt a question before checking the method.

Choose between completing the square and the discriminant, show the intervals for an inequality, and form a new quadratic for a line–curve intersection.

Questions students ask

When should I complete the square instead of using the discriminant?

Complete the square for a turning point, an extreme value or a range of y-values. Use the discriminant to count roots or find a condition on a parameter. For “always positive”, either method works if you justify the strict inequality.

Does D = 0 mean there are no roots?

No. It means two equal real roots, corresponding to one distinct x-value. The graph touches the x-axis. The square root in the quadratic formula is zero, so the plus and minus choices give the same answer.

Does a negative minimum always mean there are two roots?

For an upward-opening quadratic over all real x, yes. Its lowest point is below the axis and both arms rise, so there are two crossings. A maximum needs separate reasoning. A downward-opening curve with a negative maximum has no real roots.

Can I divide an inequality by a factor containing x?

Only after accounting for where that factor is positive, negative or zero. Its sign can change, which changes the inequality direction, and division can lose boundary values. A sign diagram usually makes these cases clearer.