2B, 2C - Quadratic roots and the discriminant

A quadratic can have two different real solutions, a repeated real solution or no real solutions. In 2B you will use the discriminant to distinguish these cases and determine conditions on coefficients; in 2C you will find the roots exactly with the quadratic formula, ready to use expressions involving them.

The quadratic formula

Does $x^2-4x+5=0$ have a real solution, or has a calculator missed one?

You already know that completing the square turns its left side into $(x-2)^2+1$ . This is at least $1$ , so it cannot equal zero for real $x$ . We will develop a formula that both finds roots when they exist and explains this failure.

Root

A root of an equation is a value of the variable that makes the equation true. A real root of $ax^2+bx+c=0$ is also the $x$ -coordinate of a point where the graph $y=ax^2+bx+c$ meets the $x$ -axis.

Completing a square produces one method for every quadratic. Starting with $ax^2+bx+c=0$ , where $a\ne0$ , multiply by $4a$ , then add the term needed for a square:

\begin{aligned} 4a^2x^2+4abx+4ac&=0 &&\text{multiply the equation by }4a\\ 4a^2x^2+4abx+b^2&=b^2-4ac &&\text{add }b^2\text{ and subtract }4ac\\ (2ax+b)^2&=b^2-4ac &&\text{recognise the square} \end{aligned}

When the right side is nonnegative, taking both signs of its square root and rearranging gives the quadratic formula.

Quadratic formula

For real coefficients with $a\ne0$ , the real roots of $ax^2+bx+c=0$ , when $b^2-4ac\ge0$ , are

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

The sign $\pm$ means evaluate once with $+$ and once with $-$ . If the square root is zero, these give the same value.

Not on the formulae sheet: learn it.

Put the equation in the form $ax^2+bx+c=0$ before identifying the coefficients. Keep the signs attached to $b$ and $c$ , and simplify the root before rounding anything.

Worked example 1: exact irrational roots

Question. Solve $x^2-4x+1=0$ , giving exact answers.

Plan. Identify $a=1$ , $b=-4$ , $c=1$ , then substitute into the formula and simplify the surd.

\begin{aligned} x&=\frac{-(-4)\pm\sqrt{(-4)^2-4(1)(1)}}{2(1)} &&\text{substitute signed coefficients}\\ &=\frac{4\pm\sqrt{12}}{2} &&\text{evaluate inside the root}\\ &=\frac{4\pm2\sqrt3}{2} &&\text{extract the square factor }4\\ &=2\pm\sqrt3 &&\text{divide both numerator terms by }2 \end{aligned}

Answer. $x=2+\sqrt3$ or $x=2-\sqrt3$ .

Check. The equation is also $(x-2)^2-3=0$ . Substituting either answer gives $(\pm\sqrt3)^2-3=0$ , so both values satisfy it exactly.

Change only the leading coefficient from $1$ to $2$ . The denominator changes too; simplifying a surd does not remove that denominator.

Worked example 2: a nonunit leading coefficient

Question. Solve $2x^2-4x+1=0$ , giving exact answers.

Plan. Use $a=2$ , $b=-4$ , $c=1$ , retaining the denominator $2a=4$ .

\begin{aligned} x&=\frac{4\pm\sqrt{16-4(2)(1)}}{4} &&\text{substitute into the formula}\\ &=\frac{4\pm\sqrt8}{4} &&\text{calculate the radicand}\\ &=\frac{4\pm2\sqrt2}{4} &&\text{simplify the surd}\\ &=1\pm\frac{\sqrt2}{2} &&\text{divide each term by }4 \end{aligned}

Answer. $x=1+\frac{\sqrt2}{2}$ or $x=1-\frac{\sqrt2}{2}$ .

Check. Completing the square gives $2(x-1)^2-1=0$ . Each answer makes the left side $2\left(\frac{\sqrt2}{2}\right)^2-1=2\times\frac12-1=0$ .

Two sign mistakes can change a correct formula into the wrong calculation.

Watch out

If $b=-4$ , then $-b=4$ and $b^2=(-4)^2=16$ . Writing $-4^2=-16$ has lost the brackets and is not the square of $b$ . The denominator $2a$ divides the whole numerator, including both terms.

The next roots also remain exact after simplifying a square factor.

Counting real roots

The number inside the formula's square root tells you how many different real roots are possible.

Compare three equations differing only in their constant term:

\begin{aligned} x^2-4x+3&=(x-2)^2-1 &&\text{zero at }x=1,3\\ x^2-4x+4&=(x-2)^2 &&\text{zero only at }x=2\\ x^2-4x+5&=(x-2)^2+1 &&\text{never zero for real }x \end{aligned}

Their graphs show a curve crossing the axis twice, touching it once or missing it entirely.

[DIAGRAM: asset_name: Three discriminants and three root patterns; asset_slug: discriminant-root-cases; caption: Changing the constant gives two unequal real roots, equal real roots or no real roots; description: Three coordinate plots on identical scales show y equals x squared minus 4x plus 3, plus 4 and plus 5. Their discriminants are respectively 4, 0 and minus 4. The first crosses the x-axis at 1 and 3, the second touches at 2, and the third has minimum 1 at x equals 2 and never meets the x-axis.]
Changing the constant gives two unequal real roots, equal real roots or no real roots

The algebra explains all three pictures.

Discriminant

The discriminant of $ax^2+bx+c=0$ , where $a\ne0$ , is

D=b^2-4ac.

It is the expression under the square root in the quadratic formula.

If $D>0$ , then $\sqrt D>0$ : adding and subtracting it gives two unequal real roots.
If $D=0$ , the formula gives $x=-\frac b{2a}$ twice: equal real roots, also called a repeated root. There is one distinct value, and the curve touches the axis there.
If $D<0$ , no real number has square $D$ : there are no real roots, and the curve does not meet the axis.

You do not need to calculate the roots just to classify them.

Worked example 3: equal real roots

Question. Determine the nature of the roots of $x^2-4x+4=0$ .

Plan. Calculate the discriminant and interpret its sign.

\begin{aligned} D&=(-4)^2-4(1)(4) &&\text{use }a=1,\ b=-4,\ c=4\\ &=16-16=0 &&\text{compare the two terms} \end{aligned}

Answer. The roots are equal and real because $D=0$ .

Check. Factorising gives $(x-2)^2=0$ . Both factors give $x=2$ , one distinct repeated root.

Increase only the constant from $4$ to $5$ . The discriminant moves below zero, so the repeated root disappears from the real number line.

Worked example 4: no real roots

Question. Determine the nature of the roots of $x^2-4x+5=0$ .

Plan. Evaluate the discriminant; a negative value rules out real roots.

\begin{aligned} D&=(-4)^2-4(1)(5) &&\text{keep the negative coefficient in brackets}\\ &=16-20=-4 &&\text{calculate exactly}\\ D&<0 &&\text{interpret the sign} \end{aligned}

Answer. There are no real roots.

Check. The left side is $(x-2)^2+1\ge1$ , so it cannot be zero for real $x$ . This independently confirms the conclusion.

A conclusion matters as much as the calculation that supports it.

Exam tip

Write what the sign means: for example, “ $D=-4<0$ , so there are no real roots.” A calculator error message alone does not explain why a real root is impossible.

Classify the next equation without calculating its roots.

Conditions on coefficients

A coefficient containing an unknown constant turns root classification into a condition on that constant.

A parameter is a constant whose value may vary between equations. For instance, changing $k$ in $x^2-6x+k=0$ gives a family of quadratics. Calculate the discriminant in terms of the parameter, then use $D>0$ , $D=0$ or $D<0$ according to the roots requested.

The boundary $D=0$ belongs to equal roots. Strict inequalities are needed for unequal roots and for no real roots. Also check the leading coefficient: if it becomes zero, the equation is no longer quadratic and the discriminant rule does not classify it.

Worked example 5: a parameter in the constant term

Question. For $x^2-6x+k=0$ , find the values of $k$ giving equal real roots, unequal real roots and no real roots.

Plan. Express the discriminant in terms of $k$ , then solve each sign condition.

\begin{aligned} D&=(-6)^2-4(1)(k)=36-4k &&\text{identify the coefficients}\\ D=0&\iff k=9 &&\text{the boundary case}\\ D>0&\iff 36>4k\iff k<9 &&\text{two unequal real roots}\\ D<0&\iff 36<4k\iff k>9 &&\text{no real roots} \end{aligned}

Answer. Equal real roots: $k=9$ . Unequal real roots: $k<9$ . No real roots: $k>9$ .

Check. At the boundary, the expression is $(x-3)^2$ . On either side, $k=8$ gives $(x-2)(x-4)$ , with two different roots, while $k=10$ gives $(x-3)^2+1$ , which is never zero. The equivalent equation $(x-3)^2=9-k$ confirms all three conditions.

Now move the parameter from the constant term to the leading coefficient, leaving the other two coefficients as $-6$ and $1$ . The discriminant happens to be the same, but $k=0$ now needs separate treatment.

Worked example 6: checking when the equation stops being quadratic

Question. Classify the roots of $kx^2-6x+1=0$ in terms of $k$ . Deal separately with $k=0$ .

Plan. Apply the discriminant only when $k\ne0$ , then substitute $k=0$ into the original equation.

For $k\ne0$ ,

\begin{aligned} D&=(-6)^2-4(k)(1)=36-4k &&\text{now }a=k\\ D=0&\iff k=9 &&\text{equal real roots}\\ D>0&\iff k<9 &&\text{retain the condition }k\ne0\\ D<0&\iff k>9 &&\text{no real roots} \end{aligned}

At $k=0$ , the original equation becomes $-6x+1=0$ , so $x=\frac16$ . It is linear, not a quadratic with repeated roots.

Answer. Equal real roots for $k=9$ ; unequal real roots for $k<9$ with $k\ne0$ ; no real roots for $k>9$ . At $k=0$ , there is the single linear solution $x=\frac16$ .

Check. At $k=9$ , the left side is $(3x-1)^2$ . At $k=8$ , it is $(4x-1)(2x-1)$ , giving different roots $\frac14$ and $\frac12$ . At $k=10$ , it is $10(x-\frac3{10})^2+\frac1{10}>0$ . Substitution of $k=0$ , $x=\frac16$ gives $-1+1=0$ .

The restriction comes from the equation's degree, not from solving the inequality.

Watch out

For $kx^2-6x+1=0$ , the inequality $D>0$ alone would include $k=0$ . Exclude it before claiming two quadratic roots. The formula assumes a nonzero leading coefficient.

The same check applies whenever a parameter can remove the squared term.

Quick Recap

Put a quadratic in the form $ax^2+bx+c=0$ , with $a\ne0$ , before using the formula.
Keep both signs and simplify exact surds; divide the whole numerator by $2a$ .
The sign of $D=b^2-4ac$ distinguishes unequal real roots, equal real roots and no real roots.
For parameter conditions, justify strict endpoints and check whether the leading coefficient can vanish.

The opening equation has $D=-4$ , so it has no real solution. Its completed square $(x-2)^2+1$ explains the same result geometrically.

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2B, 2C - Quadratic roots and the discriminant

The quadratic formula

Counting real roots

Conditions on coefficients

Quick Recap

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