Math · College algebra · Worked example

Count real solutions with the discriminant

Without solving, decide how many real solutions each equation has: $x^{2} - 6 x + 9 = 0$ , $x^{2} + 2 x + 5 = 0$ and $2 x^{2} - 3 x - 4 = 0$ . Then solve the one with exactly one solution.

x^{2} - 6 x + 9 = 0

Read $a$ , $b$ and $c$ for each

All three are in standard form. For the first, $a = 1$ , $b = - 6$ and $c = 9$ . For the second, $a = 1$ , $b = 2$ and $c = 5$ . For the third, $a = 2$ , $b = - 3$ and $c = - 4$ .

Work out each discriminant

The first gives $36 - 36 = 0$ . The second gives $4 - 20 = - 16$ . The third gives $9 + 32 = 41$ . Watch the signs, since $- 4 a c$ with a negative $c$ becomes a plus.

\begin{aligned} {(- 6)}^{2} - 4 (1) (9) & = 0 \\ 2^{2} - 4 (1) (5) & = - 16 \\ {(- 3)}^{2} - 4 (2) (- 4) & = 41 \end{aligned}

Read the count

A discriminant of $0$ means one repeated real solution. $- 16$ is negative, so the second equation has no real solution. $41$ is positive, so the third has two. $41$ is not a perfect square, so those two are irrational.

Solve the one with $D = 0$

$x^{2} - 6 x + 9$ is a perfect square, ${(x - 3)}^{2}$ . Setting it to zero gives $x - 3 = 0$ , so $x = 3$ . The formula agrees, since $- b / (2 a) = 6 / 2 = 3$ .

\begin{array}{l} x^{2} - 6 x + 9 = 0 \\ {(x - 3)}^{2} = 0 \\ x = 3 \end{array}

Match each with its graph

The parabola $y = x^{2} - 6 x + 9$ touches the x-axis at its vertex, $(3, 0)$ , and turns back up. $y = x^{2} + 2 x + 5$ has its vertex at $(- 1, 4)$ and never reaches the axis. $y = 2 x^{2} - 3 x - 4$ crosses it twice.

Result

$x^{2} - 6 x + 9 = 0$ has one real solution, $x = 3$ . $x^{2} + 2 x + 5 = 0$ has none. $2 x^{2} - 3 x - 4 = 0$ has two.

D = 0, D = - 16, D = 41

Your turn

1. One like the example

How many real solutions does $x^{2} + 4 x + 7 = 0$ have?

Show the answer and explanation

None. $D = - 12$ .

Here $a = 1$ , $b = 4$ and $c = 7$ , so $D = 16 - 28 = - 12$ . The discriminant is negative, so there is no real solution, and the parabola $y = x^{2} + 4 x + 7$ stays above the x-axis.

\begin{array}{l} D = 4^{2} - 4 (1) (7) \\ D = - 12 \end{array}

2. A variation

Show that $9 x^{2} - 12 x + 4 = 0$ has exactly one real solution, and find it.

Show the answer and explanation

$D = 0$ , and $x = 2 / 3$ .

$D = 144 - 4 (9) (4)$ $= 144 - 144$ $= 0$ , so there is one repeated solution. The quadratic is the perfect square ${(3 x - 2)}^{2}$ , and $3 x - 2 = 0$ gives $x = 2 / 3$ , which is $- b / (2 a) = 12 / 18$ .

\begin{array}{l} 9 x^{2} - 12 x + 4 = 0 \\ {(3 x - 2)}^{2} = 0 \\ x = \frac{2}{3} \end{array}

3. Somewhere new

For what value of $k$ does $x^{2} + 6 x + k = 0$ have exactly one real solution?

Show the answer and explanation

$k = 9$ .

One solution means $D = 0$ . Here $D = 6^{2} - 4 (1) (k)$ $= 36 - 4 k$ , and $36 - 4 k = 0$ gives $k = 9$ . Check: $x^{2} + 6 x + 9 = {(x + 3)}^{2}$ , whose only zero is $x = - 3$ .

\begin{array}{l} 6^{2} - 4 k = 0 \\ 36 = 4 k \\ k = 9 \end{array}

Keep exploring

Open Graph and change the $9$ in $x^{2} - 6 x + 9$ to $8$ . The discriminant becomes $4$ , and the parabola now crosses the x-axis at $2$ and $4$ .

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Study the 6 key terms Check the repeated root in Math Compare the three parabolas in Graph Open worked example on a board Quadratic formula in Math Reference

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Read aa, bb and cc for each

Work out each discriminant

Read the count

Solve the one with D=0D=0

Match each with its graph

Result

Your turn

1. One like the example

None. D=−12D=−12.

2. A variation

D=0D=0, and x=2⁢/3x=2/3.

3. Somewhere new

k=9k=9.

Keep exploring

Study it, then try it

Read $a$ , $b$ and $c$ for each

Solve the one with $D = 0$

None. $D = - 12$ .

$D = 0$ , and $x = 2 / 3$ .

$k = 9$ .