# The Discriminant

## The Discriminant

The discriminant is defined as $$\Delta ={b}^{2}-4ac$$.

This is the expression under the square root in the quadratic formula. The discriminant determines the nature of the roots of a quadratic equation. The word ‘nature’ refers to the types of numbers the roots can be — namely real, rational, irrational or imaginary. $$Δ$$ is the Greek symbol for the letter D.

For a quadratic function $$f(x)=a{x}^{2}+bx+c$$, the solutions to the equation $$f(x)=0$$ are given by the formula

$$x=\dfrac{-b±\sqrt{{b}^{2}-4ac}}{2a}=\dfrac{-b±\sqrt{\Delta }}{2a}$$

• If $$Δ < 0$$, then roots are imaginary (non-real) and beyond the scope of this book.

• If $$Δ \geq 0$$, the expression under the square root is non-negative and therefore roots are real. For real roots, we have the following further possibilities.

• If $$Δ = 0$$, the roots are equal and we can say that there is only one root.

• If $$Δ > 0$$, the roots are unequal and there are two further possibilities.

• $$Δ$$ is the square of a rational number: the roots are rational.

• $$Δ$$ is not the square of a rational number: the roots are irrational and can be expressed in decimal or surd form.

 Nature of roots Discriminant $$a>0$$ $$a<0$$ Roots are non-real $$\Delta <0$$ Roots are real and equal $$\Delta=0$$ Roots are real and unequal:rational rootsirrational roots $$Δ > 0$$$$Δ =$$ squared rational$$Δ =$$ not squared rational

## Example

### Question

Show that the roots of $$x^2-2x-7=0$$ are irrational.

### Interpret the question

For roots to be real and irrational, we need to calculate $$Δ$$ and show that it is greater than zero and not a perfect square.

### Check that the equation is in standard form $$ax^2+bx+c=0$$

$x^2-2x-7=0$

### Identify the coefficients to substitute into the formula for the discriminant

$a = 1; \qquad b = -2; \qquad c = -7$

### Write down the formula and substitute values

\begin{align*} Δ &= b^2-4ac \\ &= (-2)^2-4(1)(-7) \\ &= 4+28 \\ &= 32 \end{align*}

We know that $$32 > 0$$ and is not a perfect square.

The graph below shows the roots of the equation $$x^2 – 2x – 7 = 0$$. Note that the graph does not form part of the answer and is included for illustration purposes only.

We have calculated that $$Δ > 0$$ and is not a perfect square, therefore we can conclude that the roots are real, unequal and irrational.

## Example

### Question

For which value(s) of $$k$$ will the roots of $$6x^2 + 6 = 4kx$$ be real and equal?

### Interpret the question

For roots to be real and equal, we need to solve for the value(s) of $$k$$ such that $$\Delta =0$$.

### Check that the equation is in standard form $$ax^2+bx+c=0$$

$6x^2-4kx+6=0$

### Identify the coefficients to substitute into the formula for the discriminant

$a = 6; \qquad b = -4k; \qquad c = 6$

### Write down the formula and substitute values

\begin{align*} Δ &= b^2-4ac \\ &= (-4k)^2-4(6)(6) \\ &= 16k^2-144 \end{align*}

For roots to be real and equal, $$\Delta =0$$.

\begin{align*} Δ &= 0 \\ 16k^2 – 144 &= 0 \\ 16(k^2-9)&= 0 \\ (k-3)(k+3)&= 0 \end{align*}

Therefore $$k = 3$$ or $$k = -3$$.

### Check both answers by substituting back into the original equation

For $$k = 3$$: \begin{align*} 6x^2-4(3)x + 6 &= 0 \\ 6x^2-12x + 6 &= 0 \\ x^2-2x + 1 &= 0 \\ (x-1)(x-1)&= 0 \\ (x-1)^2&= 0 \\ \text{Therefore } x &= 1 \end{align*}

We see that for $$k = 3$$ the quadratic equation has real, equal roots $$x = 1$$.

For $$k = – 3$$: \begin{align*} 6x^2-4(-3)x + 6 &= 0 \\ 6x^2+12x + 6 &= 0 \\ x^2+2x + 1 &= 0 \\ (x+1)(x+1)&= 0 \\ (x+1)^2&= 0 \\ \text{Therefore } x &= -1 \end{align*}

We see that for $$k = -3$$ the quadratic equation has real, equal roots $$x = -1$$.

For the roots of the quadratic equation to be real and equal, $$k = 3$$ or $$k = -3$$. 

## Example

### Question

Show that the roots of $$(x+h)(x+k)=4d^2$$ are real for all real values of $$h, k$$ and $$d$$.

### Interpret the question

For roots to be real, we need to calculate $$Δ$$ and show that $$Δ \ge 0$$ for all real values of $$h$$, $$k$$ and $$d$$.

### Check that the equation is in standard form $$ax^2+bx+c=0$$

Expand the brackets and gather like terms

\begin{align*} (x+h)(x+k) &= 4d^2 \\ x^2 + hx +kx +hk -4d^2 &= 0 \\ x^2 + (h+k)x + (hk -4d^2) &= 0 \end{align*}

### Identify the coefficients to substitute into the formula for the discriminant

$a = 1; \qquad b = h+k; \qquad c = hk-4d^2$

### Write down the formula and substitute values

\begin{align*} \Delta &= b^2-4ac \\ &= (h+k)^2-4(1)(hk-4d^2) \\ &= h^2+2hk+k^2-4hk+16d^2 \\ &= h^2-2hk+k^2+16d^2 \\ &= (h – k)^2+(4d)^2 \end{align*}

For roots to be real, $$\Delta \geq 0$$.

\begin{align*} \text{We know that } (4d)^2 &\geq 0 \\ \text{and } (h – k)^2 &\geq 0 \\ \text{so then } (h – k)^2+ (4d)^2 &\geq 0 \\ \text{therefore } \Delta &\geq 0 \end{align*}

We have shown that $$\Delta \geq 0$$, therefore the roots are real for all real values of $$h, k$$ and $$d$$.