One Number Tells You Whether a Quadratic Has Two Roots, One Root or None

In the quadratic formula $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$, everything depends on the number under the square root. That number is the discriminant:

$$D=b^2-4ac$$

Its sign tells you at once whether there are two roots, one repeated root, or no real roots at all. This part covers the nature of roots, equal-root problems, finding unknown coefficients and equations with surds or letters.

**If $D>0$ the roots are real and distinct, if $D=0$ they are real and equal, and if $D<0$ there are no real roots.

- $D>0$** — two different real roots; if $D$ is a perfect square and $a,b,c$ are rational, the roots are rational
- **$D=0$** — one repeated root, $x=-\frac{b}{2a}$
- **$D<0$ — the square root of a negative number is not real, so no real roots

Worked examples.**

$$2x^2-5x+3=0:D=25-24=1>0\text{real, distinct, rational}$$
$$4x^2-12x+9=0:D=144-144=0\text{real and equal, }x=\frac{12}{8}=\frac{3}{2}$$
$$x^2+x+1=0:D=1-4=-3<0\text{no real roots}$$

An everyday example. **A ball thrown up at a school ground has height $h=20t-5t^2$ metres.** Can it reach $25$ m? $t^2-4t+5=0$ gives $D=16-20=-4<0$, so never. For $20$ m, $t^2-4t+4=0$ gives $D=0$: it just reaches $20$ m at $t=2$ s.

The substance. **Real roots means $D\ge 0$**, which includes the equal-roots case.

**Write the discriminant in terms of $k$, set it equal to zero, and solve for $k$, rejecting any value that makes the coefficient of $x^2$ zero.

Worked example 1.** $k{x}^2-6x+1=0$ has equal roots.

$$D=36-4k=0k=9$$

Worked example 2. $x^2+(k-3)x+k=0$ has equal roots.

$$D=(k-3{)}^2-4k=k^2-10k+9=(k-1)(k-9)=0k=1\text{ or }9$$

Check: $k=1$ gives $x^2-2x+1=(x-1{)}^2$; $k=9$ gives $x^2+6x+9=(x+3{)}^2$.

Worked example 3. $(k+1)x^2-2(k-1)x+1=0$ has equal roots.

$$D=4(k-1{)}^2-4(k+1)=4(k^2-3k)=4k(k-3)=0k=0\text{ or }3$$

Neither value makes $k+1=0$, so both are accepted.

An everyday example. A ball thrown at exactly the right speed just touches the ceiling once — like one repeated root.

The boundary case. **If a value of $k$ makes the $x^2$ coefficient zero**, the equation is no longer quadratic, so that value must be rejected.

A root satisfies the equation, so substitute it and solve for the unknown coefficient.

Worked example 1. $x=2$ is a root of $x^2+kx-6=0$. Find $k$ and the other root.

$$4+2k-6=0k=1$$
$$x^2+x-6=(x+3)(x-2)=0\text{other root}=-3$$

Worked example 2. $3$ and $-\frac{1}{2}$ are roots of $a{x}^2+bx-3=0$. Find $a$ and $b$.

$$x=3:9a+3b-3=0\Rightarrow 3a+b=1$$
$$x=-\frac{1}{2}:\frac{a}{4}-\frac{b}{2}-3=0\Rightarrow a-2b=12$$

From the first, $b=1-3a$. Then $a-2+6a=12$, so $a=2$ and $b=-5$.

Check: $2x^2-5x-3=(2x+1)(x-3)$, with roots $3$ and $-\frac{1}{2}$.

An everyday example. Given a root as a hint, substitute it first.

The substance. Two given roots give two equations, enough to find two unknown coefficients.

Split the middle term using factors of the product of the first and last coefficients, even when they contain surds or letters, and then substitute each root back to verify.

Worked example 1 — surds. Solve $\sqrt{3}{x}^2+10x+7\sqrt{3}=0$.

Product $=\sqrt{3}\times 7\sqrt{3}=21$, sum $=10$: use $7$ and $3$.

$$\sqrt{3}{x}^2+7x+3x+7\sqrt{3}=x(\sqrt{3}x+7)+\sqrt{3}(\sqrt{3}x+7)=(x+\sqrt{3})(\sqrt{3}x+7)$$
$$x=-\sqrt{3}\text{or}x=-\frac{7}{\sqrt{3}}=-\frac{7\sqrt{3}}{3}$$

**Verify $x=-\sqrt{3}$:** $\sqrt{3}(3)-10\sqrt{3}+7\sqrt{3}=0$.

Worked example 2 — letters. Solve $ab{x}^2+(b^2-ac)x-bc=0$.

$$ab{x}^2+b^{2}x-acx-bc=bx(ax+b)-c(ax+b)=(bx-c)(ax+b)$$
$$x=\frac{c}{b}\text{or}x=-\frac{b}{a}$$

**Verify $x=\frac{c}{b}$:** $\frac{a{c}^2}{b}+bc-\frac{a{c}^2}{b}-bc=0$.

An everyday example. Surds appear naturally in the diagonal of a square tile.

The substance. Substitution is the reliable check when roots are messy.