Quadratic Functions & Discriminant

Secondary 3 and 4 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 → practice24 min read · practice linked below

One quadratic, three useful forms

This shared guide supports Secondary 3 learning and Secondary 4 revision.

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.

Equations, substitutions and modelling

Factorisation uses the zero-product rule: if a product is zero, at least one factor is zero. Completing the square and the quadratic formula also work when convenient rational factors do not exist. Rearrange to zero first, and preserve any restrictions from denominators or from the context.

Example 5: exact roots from the quadratic formula

Solve $2x^2-3x-1=0$ . Here $a=2,b=-3,c=-1$ , so

x=\frac{3\pm\sqrt{(-3)^2-4(2)(-1)}}{4}=\frac{3\pm\sqrt{17}}4.

The two exact roots are $(3-\sqrt{17})/4$ and $(3+\sqrt{17})/4$ . As a check, their sum is $3/2$ and product is $-1/2$ : multiplying $2(x-r)(x-s)$ therefore recovers the original polynomial. These relations follow by expanding, not by treating the signs as a memorised guess.

Example 6: a quadratic after substitution

Solve $x^4-5x^2+4=0$ for real $x$ . Put $u=x^2$ , remembering $u\ge0$ . Then

u^2-5u+4=(u-1)(u-4)=0.

Thus $x^2=1$ or $x^2=4$ , giving $x=-2,-1,1,2$ . All four values satisfy the original equation. Writing only $u=1,4$ would stop before answering the question. A negative value of $u$ would have been rejected because a real square cannot be negative.

The same idea works for a repeated expression such as $x+1/x$ , but that substitution requires $x\ne0$ , and each resulting equation must still be solved for $x$ .

Example 7: a linear and a nonlinear equation

Solve $x+y=5$ and $x^2+y^2=13$ simultaneously. Substitute $y=5-x$ into the second equation:

x^2+(5-x)^2=13\quad\Longrightarrow\quad 2x^2-10x+12=0.

Dividing by 2 and factorising gives $(x-2)(x-3)=0$ . If $x=2$ , then $y=3$ ; if $x=3$ , then $y=2$ . The solutions are the ordered pairs $(2,3)$ and $(3,2)$ . In each pair the sum is 5 and the sum of squares is 13. Substitution works because the linear equation expresses the same $y$ that must satisfy the nonlinear equation.

Example 8: a maximum with a restricted domain

A rectangle has perimeter 20 cm and one side $x$ cm. Find its largest area, then find the largest area if that side is restricted to $1\le x\le4$ .

The other side is $10-x$ , so the physical domain is $0<x<10$ . The area is

A=x(10-x)=25-(x-5)^2.

The unrestricted rectangle has maximum area $25\text{ cm}^2$ at $x=5$ , a square. Under $1\le x\le4$ , that turning point is unavailable. The square $(x-5)^2$ is smallest at $x=4$ , so the restricted maximum is $24\text{ cm}^2$ , for sides 4 cm and 6 cm. If the restriction were $1\le x<4$ , the area would approach 24 but never reach it; there would be no maximum.

Example 9: when a parameter removes the quadratic term

Find all real $k$ for which $(k-1)x^2+2x+1=0$ has at least one real solution.

If $k\ne1$ , the equation is quadratic and

D=2^2-4(k-1)=8-4k\ge0\quad\Longrightarrow\quad k\le2.

That gives $k\le2$ , with $k=1$ temporarily excluded. At $k=1$ , the original equation becomes $2x+1=0$ , which has the real solution $x=-1/2$ . Therefore the complete answer is $k\le2$ . If the wording required a quadratic equation with real roots, the answer would instead exclude $k=1$ . At $k=2$ , the repeated root $x=-1$ is allowed.

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 Secondary 3 quadratic functions and inequalities questions or Secondary 4 quadratic functions and inequalities 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.