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.

a5f116a08bfe4611c25ea2f2594864c9.png

Nature of rootsDiscriminant\(a>0\)\(a<0\)
Roots are non-real\(\Delta <0\)d484cf42f62b770f6d60b67d88685f60.png5ef954ace3ab3a47ba503f1b540f2823.png
Roots are real and equal\(\Delta=0\)1a3fbcbaaa98dc58c50d36ef7d1789e7.pngaf29fd0a2a301005ba0ed1e0a2fbd60f.png

Roots are real and unequal:

  • rational roots
  • irrational roots

\(Δ > 0\)

  • \(Δ =\) squared rational
  • \(Δ =\) not squared rational
24f6df404b10a120e91bef393e7bfde2.png592c12cf2fb349966002b4f4e78c7852.png

 

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.

8a71c3a5a2688633d3461e0e4c770d62.png

Write the final answer

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 \(x = -1\).

Write the final answer

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*}

Write the final answer

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

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