2.4 The Discriminant

What you'll learn

Decide how many real solutions a quadratic equation has without solving it fully.
Use the discriminant $b^2-4ac$ to test for two distinct, equal, or no real roots.
Find ranges of constants such as $k$ , $n$ , or $p$ .
Apply the same idea to curve-line intersections and tangents.

Prerequisite: quadratics, coefficients, and roots

Before using the discriminant, you need the quadratic written in standard form.

Definition

Quadratic equation

A quadratic equation in $x$ is an equation that can be written as $ax^2+bx+c=0$ , where $a≠0a\neq 0$ .

The numbers or algebraic expressions $a$ , $b$ , and $c$ are called coefficients.
A root is a solution for $x$ .
A real root is a solution that is a real number, so it appears on the number line.

The discriminant only works once you have identified $a$ , $b$ , and $c$ correctly.

Example

Identifying $a$ , $b$ , and $c$

The equation $5x^2+(k-3)x-2=0$ , where $k$ is a constant, is already in standard form.

Compare it with the general form:
$ax^2+bx+c=0$
Match each part carefully:
$a=5,b=k−3,c=−2a=5,\quad b=k-3,\quad c=-2$
Notice that the minus sign belongs to the constant term, so $c = - 2$ , not 2.

Common Mistake

Forgetting the sign

If the equation contains something like $- 7$ or $- (k + 1)$ , the minus sign is part of the coefficient. This is one of the most common ways to lose accuracy in discriminant questions.

The discriminant

The quadratic formula is

x=−b±b2−4ac2ax=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

The expression under the square root controls what kind of roots the quadratic has.

Definition

The discriminant

For a quadratic equation $ax^2+bx+c=0$ , the discriminant is

Δ=b2−4ac\Delta=b^2-4ac

The symbol $Δ\Delta$ is the Greek capital letter “Delta”.

Here is the graph meaning of the three possible discriminant cases.

The sign of the discriminant tells whether a quadratic graph crosses, touches, or misses the x-axis.

Three parabolas showing discriminant cases

Key Idea

What the sign of the discriminant tells you

If $Δ>0\Delta>0$ , there are two distinct real roots. Distinct means different.
If $Δ=0\Delta=0$ , there is one repeated root, often called equal roots.
If $Δ<0\Delta<0$ , there are no real roots.

Example

Deciding the number of roots

Decide how many real roots the equation $2x^2-5x+7=0$ has.

Identify the coefficients:
$a=2,b=−5,c=7a=2,\quad b=-5,\quad c=7$
Substitute into the discriminant:
$Δ=b2−4ac=(−5)2−4(2)(7)\Delta=b^2-4ac=(-5)^2-4(2)(7)$
Simplify:
$Δ=25−56=−31\Delta=25-56=-31$
Since $Δ<0\Delta<0$ , the equation has no real roots.

Using the discriminant to find possible values

A parameter is a letter that behaves like a constant in a question. For example, in $x^2+kx+6=0$ , the variable is $x$ , but $k$ is a parameter.

The usual method is:

Write down $a$ , $b$ , and $c$ .
Form $Δ=b2−4ac\Delta=b^2-4ac$ .
Use the condition from the question: greater than zero, equal to zero, or less than zero.
Solve the resulting equation or inequality.

No real roots

If a quadratic has no real roots, use $Δ<0\Delta<0$ .

Example

Finding a range for a constant

The equation $x^2+kx+6=0$ , where $k$ is a constant, has no real roots. Find the possible values of $k$ .

The solution to k^2<24 is the interval between the two boundary values on the number line.

Identify the coefficients:
$a=1,b=k,c=6a=1,\quad b=k,\quad c=6$
Since there are no real roots, use $Δ<0\Delta<0$ :
$b^2-4ac<0$
Substitute the coefficients:
$k^2-4(1)(6)<0$
Simplify:
$k^2-24<0$
Rearrange and square root both sides carefully:
$k^2<24$
Therefore the possible values are:
$−26<k<26-2\sqrt{6}<k<2\sqrt{6}$

Equal roots

If a quadratic has equal roots, use $Δ=0\Delta=0$ .

Example

Equal roots with a positive constant

The equation $kx^2+7x+k=0$ , where $k$ is a positive constant, has equal roots. Find $k$ .

Identify the coefficients:
$a=k,b=7,c=ka=k,\quad b=7,\quad c=k$
Equal roots means the discriminant is zero:
$b^2-4ac=0$
Substitute:
$7^2-4(k)(k)=0$
Simplify:
$49-4k^2=0$
Solve for $k$ :
$4k^2=49$
Take the square root:
$k=±72k=\pm\frac{7}{2}$
Since $k$ is positive, the answer is:
$k=72k=\frac{7}{2}$

Tip

Match the wording to the sign

“Two distinct real roots” means $Δ>0\Delta>0$ , “equal roots” means $Δ=0\Delta=0$ , and “no real roots” means $Δ<0\Delta<0$ .

Solving harder discriminant inequalities

Sometimes the discriminant itself becomes a quadratic expression in the parameter. You then solve a quadratic inequality.

An upward-opening quadratic expression is negative between its two roots.

Example

A quadratic inequality in the parameter

The equation $x^2+(2n-1)x+(n+2)=0$ , where $n$ is a constant, has no real roots. Find the possible values of $n$ .

Identify the coefficients:
$a=1,b=2n−1,c=n+2a=1,\quad b=2n-1,\quad c=n+2$
No real roots means $Δ<0\Delta<0$ :
$2n-1)^2-4(1)(n+2)<0$
Expand and simplify:
$(2n−1)2−4(n+2)<04n2−4n+1−4n−8<04n2−8n−7<0\begin{aligned} (2n-1)^2-4(n+2)&<0\\ 4n^2-4n+1-4n-8&<0\\ 4n^2-8n-7&<0 \end{aligned}$
Solve the boundary equation:
$4n^2-8n-7=0$
Use the quadratic formula:
$n=8±64+1128=1±112n=\frac{8\pm\sqrt{64+112}}{8}=1\pm\frac{\sqrt{11}}{2}$
Since $4n^2-8n-7$ is an upward-opening quadratic, it is negative between its roots:
$1−112<n<1+1121-\frac{\sqrt{11}}{2}<n<1+\frac{\sqrt{11}}{2}$

When the coefficient of $x^2$ depends on the parameter

Be careful if the coefficient of $x^2$ contains the parameter. If that coefficient becomes zero, the equation is no longer quadratic.

Common Mistake

Check the equation is still quadratic

The discriminant test assumes $a≠0a\neq 0$ . If $a = 0$ , the equation becomes linear, so it cannot have two distinct roots.

Example

Excluding a value that makes the equation linear

The equation $r+2)x^2+5x+r=0$ , where $r$ is a constant, has two distinct real roots. Find the possible values of $r$ .

The allowed interval from the discriminant must be split because r=-2 makes the equation linear, not quadratic.

The coefficient of $x^2$ is $r + 2$ , so the equation is quadratic only if:
$r+2≠0r+2\neq 0$
Therefore:
$r≠−2r\neq -2$
For two distinct real roots, use $Δ>0\Delta>0$ :
$5^2-4(r+2)(r)>0$
Expand and simplify:
$25-4r^2-8r>0$
Rearrange into a more standard inequality:
$4r^2+8r-25<0$
Solve the boundary equation:
$4r^2+8r-25=0$
The boundary values are:
$r=−8±64+4008=−2±292r=\frac{-8\pm\sqrt{64+400}}{8}=\frac{-2\pm\sqrt{29}}{2}$
The expression is negative between the two boundary values, but $r = - 2$ must be excluded:
$−2−292<r<−2or−2<r<−2+292\frac{-2-\sqrt{29}}{2}<r<-2 \quad\text{or}\quad -2<r<\frac{-2+\sqrt{29}}{2}$

Intersections of curves and lines

When a curve and a line intersect, their $y$ -values are equal at the points of intersection. So you set the equations equal and form an equation in $x$ .

Intersections correspond to real solutions of the equation formed by setting the curve and line equal.

No intersection means no real solutions, so use $Δ<0\Delta<0$ .
A tangent means one repeated solution, so use $Δ=0\Delta=0$ .
Two intersections means two distinct real solutions, so use $Δ>0\Delta>0$ .

Example

A curve and a line that do not meet

The curve $y=qx^2-2qx-3q$ does not intersect the line $y = 2 x - 8$ . Find the possible values of $q$ .

Set the two expressions for $y$ equal:
$qx^2-2qx-3q=2x-8$
Rearrange into a quadratic equation in $x$ :
$qx^2+(-2q-2)x+(8-3q)=0$
For no intersection, use $Δ<0\Delta<0$ :
$2q-2)^2-4(q)(8-3q)<0$
Simplify:
$4(q+1)2−4q(8−3q)<04q2+8q+4−32q+12q2<016q2−24q+4<0\begin{aligned} 4(q+1)^2-4q(8-3q)&<0\\ 4q^2+8q+4-32q+12q^2&<0\\ 16q^2-24q+4&<0 \end{aligned}$
Divide by 4:
$4q^2-6q+1<0$
Solve the boundary equation:
$q=6±36−168=3±54q=\frac{6\pm\sqrt{36-16}}{8}=\frac{3\pm\sqrt{5}}{4}$
Since the quadratic in $q$ opens upwards, the possible values are:
$3−54<q<3+54\frac{3-\sqrt{5}}{4}<q<\frac{3+\sqrt{5}}{4}$

Tangents to circles

A tangent is a straight line that touches a curve at exactly one point. For a line and circle, substituting the line equation into the circle equation gives a quadratic in $x$ . If the line is tangent, that quadratic has equal roots.

A tangent to a circle meets it at exactly one point, giving a repeated solution after substitution.

Circle with two tangent lines

Example

Finding tangent gradients exactly

The line $y = m x + 3$ is a tangent to the circle $x^2+y^2=5$ . Find the two possible values of $m$ .

Substitute $y = m x + 3$ into the circle equation:
$x^2+(mx+3)^2=5$
Expand and rearrange:
$x^2+m^2x^2+6mx+9=5$
Write it as a quadratic in $x$ :
$1+m^2)x^2+6mx+4=0$
A tangent gives equal roots, so set the discriminant equal to zero:
$6m)^2-4(1+m^2)(4)=0$
Simplify:
$36m^2-16-16m^2=0$
Solve:
$20m^2=16$
Therefore:
$m=±45=±255m=\pm\sqrt{\frac{4}{5}}=\pm\frac{2\sqrt{5}}{5}$

Exam technique

In the exam

First decide which discriminant condition the wording needs: greater than zero, equal to zero, or less than zero.
Write $a$ , $b$ , and $c$ explicitly before substituting, especially when there are brackets or parameters.
If the coefficient of $x^2$ contains a parameter, check whether any value makes the equation stop being quadratic.

Self review

Check yourself

What discriminant condition tells you that a quadratic has two distinct real roots?
If $a = k - 1$ , what value of $k$ would make the equation no longer quadratic?
Why does a tangent line lead to $Δ=0\Delta=0$ after substitution?

2.4 The Discriminant

What you'll learn

Decide how many real solutions a quadratic equation has without solving it fully.
Use the discriminant $b^2-4ac$ to test for two distinct, equal, or no real roots.
Find ranges of constants such as $k$ , $n$ , or $p$ .
Apply the same idea to curve-line intersections and tangents.

Prerequisite: quadratics, coefficients, and roots

Before using the discriminant, you need the quadratic written in standard form.

Definition

Quadratic equation

A quadratic equation in $x$ is an equation that can be written as $ax^2+bx+c=0$ , where $a≠0a\neq 0$ .

The numbers or algebraic expressions $a$ , $b$ , and $c$ are called coefficients.
A root is a solution for $x$ .
A real root is a solution that is a real number, so it appears on the number line.

The discriminant only works once you have identified $a$ , $b$ , and $c$ correctly.

Example

Identifying $a$ , $b$ , and $c$

The equation $5x^2+(k-3)x-2=0$ , where $k$ is a constant, is already in standard form.

Compare it with the general form:
$ax^2+bx+c=0$
Match each part carefully:
$a=5,b=k−3,c=−2a=5,\quad b=k-3,\quad c=-2$
Notice that the minus sign belongs to the constant term, so $c = - 2$ , not 2.

Common Mistake

Forgetting the sign

If the equation contains something like $- 7$ or $- (k + 1)$ , the minus sign is part of the coefficient. This is one of the most common ways to lose accuracy in discriminant questions.

The discriminant

The quadratic formula is

x=−b±b2−4ac2ax=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

The expression under the square root controls what kind of roots the quadratic has.

Definition

The discriminant

For a quadratic equation $ax^2+bx+c=0$ , the discriminant is

Δ=b2−4ac\Delta=b^2-4ac

The symbol $Δ\Delta$ is the Greek capital letter “Delta”.

Here is the graph meaning of the three possible discriminant cases.

The sign of the discriminant tells whether a quadratic graph crosses, touches, or misses the x-axis.

Three parabolas showing discriminant cases

Key Idea

What the sign of the discriminant tells you

If $Δ>0\Delta>0$ , there are two distinct real roots. Distinct means different.
If $Δ=0\Delta=0$ , there is one repeated root, often called equal roots.
If $Δ<0\Delta<0$ , there are no real roots.

Example

Deciding the number of roots

Decide how many real roots the equation $2x^2-5x+7=0$ has.

Identify the coefficients:
$a=2,b=−5,c=7a=2,\quad b=-5,\quad c=7$
Substitute into the discriminant:
$Δ=b2−4ac=(−5)2−4(2)(7)\Delta=b^2-4ac=(-5)^2-4(2)(7)$
Simplify:
$Δ=25−56=−31\Delta=25-56=-31$
Since $Δ<0\Delta<0$ , the equation has no real roots.

Using the discriminant to find possible values

A parameter is a letter that behaves like a constant in a question. For example, in $x^2+kx+6=0$ , the variable is $x$ , but $k$ is a parameter.

The usual method is:

Write down $a$ , $b$ , and $c$ .
Form $Δ=b2−4ac\Delta=b^2-4ac$ .
Use the condition from the question: greater than zero, equal to zero, or less than zero.
Solve the resulting equation or inequality.

No real roots

If a quadratic has no real roots, use $Δ<0\Delta<0$ .

Example

Finding a range for a constant

The equation $x^2+kx+6=0$ , where $k$ is a constant, has no real roots. Find the possible values of $k$ .

The solution to k^2<24 is the interval between the two boundary values on the number line.

Identify the coefficients:
$a=1,b=k,c=6a=1,\quad b=k,\quad c=6$
Since there are no real roots, use $Δ<0\Delta<0$ :
$b^2-4ac<0$
Substitute the coefficients:
$k^2-4(1)(6)<0$
Simplify:
$k^2-24<0$
Rearrange and square root both sides carefully:
$k^2<24$
Therefore the possible values are:
$−26<k<26-2\sqrt{6}<k<2\sqrt{6}$

Equal roots

If a quadratic has equal roots, use $Δ=0\Delta=0$ .

Example

Equal roots with a positive constant

The equation $kx^2+7x+k=0$ , where $k$ is a positive constant, has equal roots. Find $k$ .

Identify the coefficients:
$a=k,b=7,c=ka=k,\quad b=7,\quad c=k$
Equal roots means the discriminant is zero:
$b^2-4ac=0$
Substitute:
$7^2-4(k)(k)=0$
Simplify:
$49-4k^2=0$
Solve for $k$ :
$4k^2=49$
Take the square root:
$k=±72k=\pm\frac{7}{2}$
Since $k$ is positive, the answer is:
$k=72k=\frac{7}{2}$

Tip

Match the wording to the sign

“Two distinct real roots” means $Δ>0\Delta>0$ , “equal roots” means $Δ=0\Delta=0$ , and “no real roots” means $Δ<0\Delta<0$ .

Solving harder discriminant inequalities

Sometimes the discriminant itself becomes a quadratic expression in the parameter. You then solve a quadratic inequality.

An upward-opening quadratic expression is negative between its two roots.

Example

A quadratic inequality in the parameter

The equation $x^2+(2n-1)x+(n+2)=0$ , where $n$ is a constant, has no real roots. Find the possible values of $n$ .

Identify the coefficients:
$a=1,b=2n−1,c=n+2a=1,\quad b=2n-1,\quad c=n+2$
No real roots means $Δ<0\Delta<0$ :
$2n-1)^2-4(1)(n+2)<0$
Expand and simplify:
$(2n−1)2−4(n+2)<04n2−4n+1−4n−8<04n2−8n−7<0\begin{aligned} (2n-1)^2-4(n+2)&<0\\ 4n^2-4n+1-4n-8&<0\\ 4n^2-8n-7&<0 \end{aligned}$
Solve the boundary equation:
$4n^2-8n-7=0$
Use the quadratic formula:
$n=8±64+1128=1±112n=\frac{8\pm\sqrt{64+112}}{8}=1\pm\frac{\sqrt{11}}{2}$
Since $4n^2-8n-7$ is an upward-opening quadratic, it is negative between its roots:
$1−112<n<1+1121-\frac{\sqrt{11}}{2}<n<1+\frac{\sqrt{11}}{2}$

When the coefficient of $x^2$ depends on the parameter

Be careful if the coefficient of $x^2$ contains the parameter. If that coefficient becomes zero, the equation is no longer quadratic.

Common Mistake

Check the equation is still quadratic

The discriminant test assumes $a≠0a\neq 0$ . If $a = 0$ , the equation becomes linear, so it cannot have two distinct roots.

Example

Excluding a value that makes the equation linear

The equation $r+2)x^2+5x+r=0$ , where $r$ is a constant, has two distinct real roots. Find the possible values of $r$ .

The allowed interval from the discriminant must be split because r=-2 makes the equation linear, not quadratic.

The coefficient of $x^2$ is $r + 2$ , so the equation is quadratic only if:
$r+2≠0r+2\neq 0$
Therefore:
$r≠−2r\neq -2$
For two distinct real roots, use $Δ>0\Delta>0$ :
$5^2-4(r+2)(r)>0$
Expand and simplify:
$25-4r^2-8r>0$
Rearrange into a more standard inequality:
$4r^2+8r-25<0$
Solve the boundary equation:
$4r^2+8r-25=0$
The boundary values are:
$r=−8±64+4008=−2±292r=\frac{-8\pm\sqrt{64+400}}{8}=\frac{-2\pm\sqrt{29}}{2}$
The expression is negative between the two boundary values, but $r = - 2$ must be excluded:
$−2−292<r<−2or−2<r<−2+292\frac{-2-\sqrt{29}}{2}<r<-2 \quad\text{or}\quad -2<r<\frac{-2+\sqrt{29}}{2}$

Intersections of curves and lines

When a curve and a line intersect, their $y$ -values are equal at the points of intersection. So you set the equations equal and form an equation in $x$ .

Intersections correspond to real solutions of the equation formed by setting the curve and line equal.

No intersection means no real solutions, so use $Δ<0\Delta<0$ .
A tangent means one repeated solution, so use $Δ=0\Delta=0$ .
Two intersections means two distinct real solutions, so use $Δ>0\Delta>0$ .

Example

A curve and a line that do not meet

The curve $y=qx^2-2qx-3q$ does not intersect the line $y = 2 x - 8$ . Find the possible values of $q$ .

Set the two expressions for $y$ equal:
$qx^2-2qx-3q=2x-8$
Rearrange into a quadratic equation in $x$ :
$qx^2+(-2q-2)x+(8-3q)=0$
For no intersection, use $Δ<0\Delta<0$ :
$2q-2)^2-4(q)(8-3q)<0$
Simplify:
$4(q+1)2−4q(8−3q)<04q2+8q+4−32q+12q2<016q2−24q+4<0\begin{aligned} 4(q+1)^2-4q(8-3q)&<0\\ 4q^2+8q+4-32q+12q^2&<0\\ 16q^2-24q+4&<0 \end{aligned}$
Divide by 4:
$4q^2-6q+1<0$
Solve the boundary equation:
$q=6±36−168=3±54q=\frac{6\pm\sqrt{36-16}}{8}=\frac{3\pm\sqrt{5}}{4}$
Since the quadratic in $q$ opens upwards, the possible values are:
$3−54<q<3+54\frac{3-\sqrt{5}}{4}<q<\frac{3+\sqrt{5}}{4}$

Tangents to circles

A tangent to a circle meets it at exactly one point, giving a repeated solution after substitution.

Circle with two tangent lines

Example

Finding tangent gradients exactly

The line $y = m x + 3$ is a tangent to the circle $x^2+y^2=5$ . Find the two possible values of $m$ .

Substitute $y = m x + 3$ into the circle equation:
$x^2+(mx+3)^2=5$
Expand and rearrange:
$x^2+m^2x^2+6mx+9=5$
Write it as a quadratic in $x$ :
$1+m^2)x^2+6mx+4=0$
A tangent gives equal roots, so set the discriminant equal to zero:
$6m)^2-4(1+m^2)(4)=0$
Simplify:
$36m^2-16-16m^2=0$
Solve:
$20m^2=16$
Therefore:
$m=±45=±255m=\pm\sqrt{\frac{4}{5}}=\pm\frac{2\sqrt{5}}{5}$

Exam technique

In the exam

First decide which discriminant condition the wording needs: greater than zero, equal to zero, or less than zero.
Write $a$ , $b$ , and $c$ explicitly before substituting, especially when there are brackets or parameters.
If the coefficient of $x^2$ contains a parameter, check whether any value makes the equation stop being quadratic.

Self review

Check yourself

What discriminant condition tells you that a quadratic has two distinct real roots?
If $a = k - 1$ , what value of $k$ would make the equation no longer quadratic?
Why does a tangent line lead to $Δ=0\Delta=0$ after substitution?

2.4 The Discriminant

What you'll learn

Prerequisite: quadratics, coefficients, and roots

The discriminant

Using the discriminant to find possible values

No real roots

Equal roots

Solving harder discriminant inequalities

When the coefficient of $x^2$ depends on the parameter

Intersections of curves and lines

Tangents to circles

Recap questions

2.4 The Discriminant

What you'll learn

Prerequisite: quadratics, coefficients, and roots

The discriminant

Using the discriminant to find possible values

No real roots

Equal roots

Solving harder discriminant inequalities

When the coefficient of $x^2$ depends on the parameter

Intersections of curves and lines

Tangents to circles

Recap questions

2.4 The Discriminant

What you'll learn

Prerequisite: quadratics, coefficients, and roots

The discriminant

Using the discriminant to find possible values

No real roots

Equal roots

Solving harder discriminant inequalities

When the coefficient of x2x^2x2 depends on the parameter

Intersections of curves and lines

Tangents to circles

Recap questions

2.4 The Discriminant Revision Guide

2.4 The Discriminant

What you'll learn

Prerequisite: quadratics, coefficients, and roots

The discriminant

Using the discriminant to find possible values

No real roots

Equal roots

Solving harder discriminant inequalities

When the coefficient of x2x^2x2 depends on the parameter

Intersections of curves and lines

Tangents to circles

Recap questions

2.4 The Discriminant Revision Guide

When the coefficient of $x^2$ depends on the parameter

When the coefficient of $x^2$ depends on the parameter