Quadratic Formula and Nature of Roots (Discriminant), Class 10

Why the for­mula is true#

The for­mula comes from com­plet­ing the square. Since $a \neq 0$, we may fac­tor out $a$ and add and sub­tract $\displaystyle \left(\frac{b}{2a}\right)^2$:

$$\displaystyle \begin{aligned} ax^2 + bx + c &= a\left[x^2 + \frac{b}{a}x + \frac{c}{a}\right] \\ &= a\left[\left(x + \frac{b}{2a}\right)^2 - \frac{b^2}{4a^2} + \frac{c}{a}\right] \\ &= a\left[\left(x + \frac{b}{2a}\right)^2 - \frac{b^2 - 4ac}{4a^2}\right] \end{aligned}$$

So $ax^2 + bx + c = 0$ exactly when

$$\displaystyle \left(x + \frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a^2}$$

The left-hand side is a square, so it is never neg­a­tive, and $4a^2 \gt 0$. If $b^2 - 4ac \geq 0$, take square roots:

$$\displaystyle x + \frac{b}{2a} = \pm\frac{\sqrt{b^2 - 4ac}}{2a} \implies x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

If $b^2 - 4ac \lt 0$, the right-hand side is neg­a­tive and no real $x$ can make a square equal to it, so there are no real roots.

The dis­crim­i­nant and the nature of roots#

Def­i­n­i­tion#

The num­ber $D = b^2 - 4ac$ is called the dis­crim­i­nant of the qua­dratic equa­tion $ax^2 + bx + c = 0$, because it dis­crim­i­nates (tells apart) the three pos­si­ble kinds of roots.

The three cases#

Case 1: $D \gt 0$. Then $\sqrt{D}$ is a pos­i­tive real num­ber, so adding it to $-b$ and sub­tract­ing it from $-b$ give dif­fer­ent val­ues. The equa­tion has two dis­tinct real roots, $\displaystyle \frac{-b + \sqrt{D}}{2a}$ and $\displaystyle \frac{-b - \sqrt{D}}{2a}$.

Case 2: $D = 0$. Then $\sqrt{D} = 0$, and both roots equal

$$\displaystyle x = -\frac{b}{2a}$$

The equa­tion has two equal real roots. This is the repeated-root sit­u­a­tion you met in the surd exam­ple of the fac­tori­sa­tion les­son.

Case 3: $D \lt 0$. No real num­ber has a neg­a­tive square, so $\sqrt{D}$ is not a real num­ber. The equa­tion has no real roots.

Sum­mary#

A qua­dratic equa­tion $ax^2 + bx + c = 0$ has

• two dis­tinct real roots, if $b^2 - 4ac \gt 0$;
• two equal real roots, if $b^2 - 4ac = 0$;
• no real roots, if $b^2 - 4ac \lt 0$.

Method: using the dis­crim­i­nant and the for­mula#

1. Write the equa­tion in stan­dard form and read off $a$, $b$ and $c$, with their signs.
2. Com­pute $D = b^2 - 4ac$, putting neg­a­tive val­ues in brack­ets.
3. Decide the nature of the roots from the sign of $D$.
4. If $D \gt 0$, use $\displaystyle x = \frac{-b \pm \sqrt{D}}{2a}$ and sim­plify any surd.
5. If $D = 0$, the equal roots are $\displaystyle x = -\frac{b}{2a}$.
6. If $D \lt 0$, state that there are no real roots; in a word prob­lem, the sit­u­a­tion is not pos­si­ble.

Worked exam­ples#

Exam­ple 1: check­ing for real roots#

Find the dis­crim­i­nant of $2x^2 - 4x + 3 = 0$ and hence find the nature of its roots.

Here $a = 2$, $b = -4$, $c = 3$.

$$D = (-4)^2 - 4(2)(3) = 16 - 24 = -8$$

Answer: since $D = -8 \lt 0$, the equa­tion has no real roots. There is no point try­ing to fac­torise it.

Exam­ple 2: solv­ing with the for­mula#

Solve $3x^2 - 5x + 2 = 0$ using the qua­dratic for­mula.

Here $a = 3$, $b = -5$, $c = 2$, so $D = (-5)^2 - 4(3)(2) = 25 - 24 = 1 \gt 0$: two dis­tinct real roots.

$$\displaystyle x = \frac{-(-5) \pm \sqrt{1}}{2 \times 3} = \frac{5 \pm 1}{6}$$

Answer: $\displaystyle x = \frac{6}{6} = 1$ or $\displaystyle x = \frac{4}{6} = \frac{2}{3}$.

Exam­ple 3: two equal roots, found quickly#

Find the dis­crim­i­nant of $\displaystyle 3x^2 - 2x + \frac{1}{3} = 0$ and find its roots if they are real.

Here $a = 3$, $b = -2$, $\displaystyle c = \frac{1}{3}$.

$$\displaystyle D = (-2)^2 - 4(3)\left(\frac{1}{3}\right) = 4 - 4 = 0$$

So the roots are real and equal, each being

$$\displaystyle x = -\frac{b}{2a} = \frac{2}{2 \times 3} = \frac{1}{3}$$

Answer: two equal real roots, $\displaystyle \frac{1}{3}$ and $\displaystyle \frac{1}{3}$.

Exam­ple 4: find­ing an unknown coef­fi­cient#

Find the val­ues of $k$ for which $x^2 - kx + 16 = 0$ has two equal roots, and find the roots.

Here $a = 1$, $b = -k$, $c = 16$. For equal roots, $D = 0$:

$$(-k)^2 - 4(1)(16) = 0 \implies k^2 = 64 \implies k = \pm 8$$

With $k = 8$ the equa­tion is $x^2 - 8x + 16 = (x - 4)^2 = 0$, with equal roots 4 and 4. With $k = -8$ it is $(x + 4)^2 = 0$, with equal roots $-4$ and $-4$.

Answer: $k = 8$ (roots 4, 4) or $k = -8$ (roots $-4$, $-4$).

Exam­ple 5: a pole on the bound­ary of a cir­cu­lar park#

Is it pos­si­ble to design a cir­cu­lar park of diam­e­ter 13 m with a pole on its bound­ary such that the dif­fer­ence of its dis­tances from two dia­met­ri­cally oppo­site fixed gates $A$ and $B$ is 7 m? If so, how far from each gate should the pole be?

Let $P$ be the pole, and let $BP = x$ m. Then $AP = (x + 7)$ m. Since $AB$ is a diam­e­ter, the angle in a semi­cir­cle is a right angle, so $\angle APB = 90^\circ$. By Pythago­ras' the­o­rem in $\triangle APB$:

$$\begin{aligned} AP^2 + BP^2 &= AB^2 \\ (x + 7)^2 + x^2 &= 13^2 \\ x^2 + 14x + 49 + x^2 &= 169 \\ 2x^2 + 14x - 120 &= 0 \\ x^2 + 7x - 60 &= 0 \end{aligned}$$

Before solv­ing, check whether real roots exist. Here $a = 1$, $b = 7$, $c = -60$:

$$D = 7^2 - 4(1)(-60) = 49 + 240 = 289 \gt 0$$

So the equa­tion has two real roots and the design is pos­si­ble. By the for­mula, with $\sqrt{289} = 17$:

$$\displaystyle x = \frac{-7 \pm 17}{2} \implies x = 5 \ \text{or} \ x = -12$$

A dis­tance can­not be neg­a­tive, so $x = 5$.

Answer: yes, it is pos­si­ble. The pole should be 5 m from gate $B$ and $5 + 7 = 12$ m from gate $A$. Check: $5^2 + 12^2 = 25 + 144 = 169 = 13^2$.

Com­mon mis­takes#

• Writ­ing $b^2$ as $-b^2$ when $b$ is neg­a­tive. With $b = -4$, $b^2 = (-4)^2 = 16$, not $-16$.
• Read­ing $a$, $b$, $c$ before the equa­tion is in stan­dard form.
• Divid­ing only $\sqrt{D}$ by $2a$ instead of the whole numer­a­tor $-b \pm \sqrt{D}$.
• For­get­ting the minus sign in $-b$: for $b = -5$, $-b = 5$.
• Say­ing "no roots" when $D \lt 0$; the cor­rect state­ment is "no real roots".
• Accept­ing $k = 0$ as an answer when it makes the coef­fi­cient of $x^2$ zero, so that the equa­tion is no longer qua­dratic.

Try these#

1. Find the nature of the roots of $2x^2 - 7x + 3 = 0$, and the roots if real. Answer: $D = 25 \gt 0$, two dis­tinct real roots, $x = 3$ and $\displaystyle x = \frac{1}{2}$.
2. Find the nature of the roots of $x^2 + 4x + 5 = 0$. Answer: $D = -4 \lt 0$, no real roots.
3. Find the nature of the roots of $4x^2 - 12x + 9 = 0$, and the roots if real. Answer: $D = 0$, two equal real roots, $\displaystyle \frac{3}{2}$ and $\displaystyle \frac{3}{2}$.
4. Solve $x^2 - 2x - 1 = 0$ using the qua­dratic for­mula. Answer: $x = 1 + \sqrt{2}$ or $x = 1 - \sqrt{2}$.
5. Find $k$ so that $x^2 - kx + 9 = 0$ has two equal roots. Answer: $k = 6$ or $k = -6$.

Key terms#

Qua­dratic for­mula

$\displaystyle x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, giv­ing the real roots of $ax^2 + bx + c = 0$ when $b^2 - 4ac \geq 0$.

Dis­crim­i­nant

The num­ber $D = b^2 - 4ac$, whose sign decides the nature of the roots.

Nature of roots

Whether the roots are real and dis­tinct, real and equal, or not real.

Dis­tinct real roots

Two dif­fer­ent real roots, occur­ring when $D \gt 0$.

Equal roots

Two iden­ti­cal real roots, each $\displaystyle -\frac{b}{2a}$, occur­ring when $D = 0$.

Com­plet­ing the square

Rewrit­ing $x^2 + px$ as $\displaystyle \left(x + \frac{p}{2}\right)^2 - \frac{p^2}{4}$; the method behind the qua­dratic for­mula.

Angle in a semi­cir­cle

The angle sub­tended by a diam­e­ter at any point of the cir­cle, which is always $90^\circ$.

Com­mon ques­tions#

Should I use fac­tori­sa­tion or the for­mula?#

Use fac­tori­sa­tion when the split is easy to see; it is faster. Use the for­mula when it is not, or when the roots involve surds. Both give the same roots.

Does $D \lt 0$ mean my work­ing is wrong?#

Not nec­es­sar­ily. Some equa­tions gen­uinely have no real roots. Recheck $a$, $b$ and $c$, and if $D$ is still neg­a­tive, the answer is "no real roots".

If $D$ is a per­fect square, what does that tell me?#

When $a$, $b$, $c$ are inte­gers and $D$ is a per­fect square, the roots are ratio­nal, so the equa­tion can also be solved by split­ting the mid­dle term.

Why is the dis­crim­i­nant use­ful in word prob­lems?#

It answers "is this pos­si­ble?" directly. A neg­a­tive dis­crim­i­nant means no real length, age or dis­tance can sat­isfy the con­di­tions.

Why are the equal roots $\displaystyle -\frac{b}{2a}$?#

When $D = 0$, the $\pm\sqrt{D}$ part of the for­mula is zero, leav­ing $\displaystyle x = \frac{-b}{2a}$ for both roots.

Ref­er­ences#

1. National Coun­cil of Edu­ca­tional Research and Train­ing. Math­e­mat­ics: Text­book for Class X. NCERT, New Delhi.
2. Hall, H. S. and Knight, S. R. Higher Alge­bra. Macmil­lan.
3. Stew­art, J., Redlin, L. and Wat­son, S. Pre­cal­cu­lus: Math­e­mat­ics for Cal­cu­lus. Cen­gage Learn­ing.