1.1 - Counting real roots | Cambridge A-Level Mathematics

You can often decide how many real solutions a quadratic has without finding either solution. You will use the discriminant, apply it to equations containing a parameter and connect the result to where graphs cross or touch. The distinction between two roots and a repeated root will matter throughout.

The discriminant

How can you tell whether $x^2-6x+c=0$ has two, one or no real solutions while $c$ is still unknown? In the quadratic formula, just one expression controls whether the two answers separate, coincide or fail to be real.

For $a\ne0$ , the formula is

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

When the square root is positive, the plus and minus signs give different numbers. When it is zero, both signs give $-b/(2a)$ . A negative number has no real square root.

Discriminant

The discriminant of $ax^2+bx+c$ is $D=b^2-4ac$ . For the quadratic equation $ax^2+bx+c=0$ , with $a\ne0$ , it distinguishes the possible numbers of real roots.

The word “real” matters. We are counting solutions on the real number line.

Discriminant and real roots

For $ax^2+bx+c=0$ , where $a\ne0$ ,

\begin{array}{c|c} D>0&\text{two distinct real roots}\\ D=0&\text{one repeated real root}\\ D<0&\text{no real roots} \end{array}

The quadratic formula containing $D$ is in the formula booklet; learn these interpretations.

Here are two members of the opening family.

Worked example 1: a positive discriminant

Question. Determine the number of real roots of $x^2-6x+5=0$ .

Plan. Calculate $D$ and use its sign.

D=(-6)^2-4(1)(5)=36-20=16>0.

Answer. There are two distinct real roots.

Check. Factorising gives $(x-1)(x-5)=0$ , with the two different roots $1$ and $5$ .

Change only the constant from $5$ to $9$ . The discriminant reaches the boundary value zero.

Worked example 2: a repeated root

Question. Determine the number of real roots of $x^2-6x+9=0$ .

Plan. Use the same calculation with the new constant.

D=(-6)^2-4(1)(9)=0.

Answer. There is one repeated real root.

Check. The equation is $(x-3)^2=0$ , so the only distinct solution is $x=3$ .

If the constant rises to $10$ , then $D=36-40=-4$ . Completing the square gives $(x-3)^2+1=0$ , confirming that no real input works. Thus the three cases come from the same family, with the constant moving its minimum through zero.

Watch out

$D=0$ means a repeated root, not no roots. Also, $b^2$ uses the signed coefficient: if $b=-6$ , then $b^2=(-6)^2=36$ .

Calculate the sign before drawing a conclusion.

Conditions on a parameter

A parameter is a constant whose value is allowed to vary between different equations. Treat it as a coefficient while calculating the discriminant, then solve the condition the question asks for.

“Real roots” includes a repeated root and therefore requires $D\ge0$ . “Two distinct real roots” requires $D>0$ . Before either test, check that the coefficient of $x^2$ is nonzero: otherwise the equation is no longer quadratic.

Worked example 3: a parameter in the constant

Question. Find the set of values of $k$ for which $x^2-4x+k=0$ has two distinct real roots.

Plan. The leading coefficient is always $1$ , so impose $D>0$ .

\begin{aligned} D&=(-4)^2-4(1)(k)=16-4k\\ 16-4k&>0 &&\text{distinct real roots}\\ k&<4 &&\text{rearrange the linear inequality} \end{aligned}

Answer. $k<4$ .

Check. $k=3$ gives $(x-1)(x-3)=0$ , two roots. The boundary $k=4$ gives $(x-2)^2=0$ , so it must be excluded.

Now move the varying coefficient to the leading term. The extra check is whether that term can disappear.

Worked example 4: a parameter in the leading coefficient

Question. Find the set of values of $k$ for which $(k-1)x^2-4x+1=0$ has two distinct real roots.

Plan. Require $k-1\ne0$ , then calculate the discriminant.

\begin{aligned} k&\ne1 &&\text{retain a quadratic equation}\\ D&=16-4(k-1)=20-4k\\ 20-4k&>0\\ k&<5 \end{aligned}

Answer. $k<5$ , with $k\ne1$ .

Check. At $k=1$ , the original is $-4x+1=0$ : only one root, despite the positive value obtained for $D$ . At $k=5$ , it is $(2x-1)^2=0$ , the excluded repeated-root boundary. At $k=2$ , $D=12>0$ and the equation remains quadratic.

If asked for a repeated root in the second example, solve $20-4k=0$ , giving $k=5$ , then check $k-1=4\ne0$ . If asked for at least one real root, the answer is $k\le5$ : the exceptional linear case $k=1$ also has a real solution. Always interpret the original question after checking the exceptional case.

Watch out

A positive value of $b^2-4ac$ does not guarantee two roots when $a=0$ . The quadratic formula divides by $2a$ , so its root-count rule requires $a\ne0$ .

Include the exceptional leading-coefficient value in your reasoning.

A graph interpretation

A root of $ax^2+bx+c=0$ is the horizontal coordinate of a point where $y=ax^2+bx+c$ meets the $x$ -axis. Two roots give two crossings; a repeated root gives one point where the parabola touches the axis; no real roots give no meeting points.

The same reasoning applies to a horizontal line at another height. To find where $y=x^2-4x+3$ meets $y=h$ , solve $x^2-4x+3=h$ . The roots now locate the meetings with that line, rather than necessarily the $x$ -axis.

[DIAGRAM: asset_name: Crossing, touching and missing; asset_slug: discriminant-intersections; caption: The same parabola has two, one or no intersections as a horizontal line moves; description: Three equal-scale plots show y equals x squared minus 4x plus 3. The lines y equals 0, y equals minus 1 and y equals minus 2 give two crossings at x equals 1 and 3, a touch at vertex 2 comma minus 1, and no intersections respectively.]
The same parabola has two, one or no intersections as a horizontal line moves

The picture makes the algebra's boundary visible. Touching here means one common point with a repeated horizontal coordinate; no differentiation is needed to find it.

Worked example 5: two intersections

Question. Determine the number of points where $y=x^2-4x+3$ meets $y=0$ .

Plan. Equate the two heights and count the roots.

x^2-4x+3=0, \qquad D=16-12=4>0.

Answer. There are two points of intersection.

Check. The roots are $x=1,3$ , giving the two points $(1,0)$ and $(3,0)$ . The vertex $(2,-1)$ lies below the line, consistent with two crossings.

Move only the horizontal line, from height $0$ to height $-1$ .

Worked example 6: a touching line

Question. Determine the number of points where $y=x^2-4x+3$ meets $y=-1$ .

Plan. Move all terms of the equality to one side before using the discriminant.

\begin{aligned} x^2-4x+3&=-1\\ x^2-4x+4&=0\\ D&=16-16=0 \end{aligned}

Answer. There is one point of contact; the line touches the parabola.

Check. $(x-2)^2=0$ gives $x=2$ . Substitution gives $y=-1$ , so the common point is exactly the vertex $(2,-1)$ .

At height $-2$ , the reduced equation would be $(x-2)^2+1=0$ , with no real solution. For the opening family $x^2-6x+c=0$ , the discriminant is $36-4c$ : two roots for $c<9$ , a repeated root for $c=9$ , and none for $c>9$ .

Key idea

Count roots of the equation obtained by equating the two graph heights. Using the discriminant of the curve's original expression alone counts its meetings with the $x$ -axis, not with an arbitrary line.

Use the appropriate equality for a different horizontal line.

Quick Recap

The discriminant is $b^2-4ac$ ; its sign counts real roots when $a\ne0$ .
A zero discriminant gives a repeated root and a touching point.
For a parameter, state the discriminant condition and check any value that removes the quadratic term.
For graph intersections, first equate the two expressions for the height.

The opening family is controlled by $36-4c$ , with $c=9$ as its repeated-root boundary.

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1.1 - Counting real roots

The discriminant

Conditions on a parameter

A graph interpretation

Quick Recap

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