Solving Quadratic Equations by Factorisation, Class 10

Once a sit­u­a­tion has been writ­ten as a qua­dratic equa­tion, the next job is to find the value of $x$: the actual breadth, the actual num­ber of toys, the actual num­ber of mar­bles. The most direct way to do this, when it works, is fac­tori­sa­tion: write the qua­dratic expres­sion as a prod­uct of two lin­ear fac­tors and set each fac­tor equal to zero. This les­son fol­lows Sec­tion 4.3 of the NCERT Class 10 text­book and fin­ishes the prayer hall and mar­bles prob­lems from the pre­vi­ous les­son.

Roots of a qua­dratic equa­tion#

What "solv­ing" means#

Take the equa­tion $2x^2 - 3x + 1 = 0$ and sub­sti­tute $x = 1$ in the left-hand side:

$$2(1)^2 - 3(1) + 1 = 2 - 3 + 1 = 0$$

The left-hand side equals the right-hand side, so $x = 1$ makes the equa­tion true. Such a value is called a root.

Def­i­n­i­tion#

A real num­ber $\alpha$ is called a root of the qua­dratic equa­tion $ax^2 + bx + c = 0$ if $a\alpha^2 + b\alpha + c = 0$. We also say that $x = \alpha$ is a solu­tion of the equa­tion, or that $\alpha$ sat­is­fies the equa­tion.

The zeroes of the qua­dratic poly­no­mial $ax^2 + bx + c$ and the roots of the equa­tion $ax^2 + bx + c = 0$ are the same num­bers. A qua­dratic poly­no­mial has at most two zeroes, so a qua­dratic equa­tion has at most two roots.

The key result: the zero-prod­uct rule#

State­ment#

If the prod­uct of two real num­bers is zero, then at least one of them is zero:

$$pq = 0 \implies p = 0 \ \text{or} \ q = 0$$

Why it is true#

Sup­pose $pq = 0$ and $p \neq 0$. Then $p$ has a rec­i­p­ro­cal $\displaystyle \frac{1}{p}$, and mul­ti­ply­ing both sides by it gives $\displaystyle q = \frac{1}{p} \times 0 = 0$. So if $p$ is not zero, $q$ must be. This is exactly why fac­tori­sa­tion solves a qua­dratic equa­tion: once $ax^2 + bx + c$ is writ­ten as $(px + q)(rx + s)$, the equa­tion holds pre­cisely when $px + q = 0$ or $rx + s = 0$.

Split­ting the mid­dle term#

You learnt in Class IX to fac­torise $ax^2 + bx + c$ by split­ting the mid­dle term: find two num­bers whose sum is $b$ and whose prod­uct is $a \times c$, write $bx$ as the sum of two terms using them, and then group.

Method: solv­ing by fac­tori­sa­tion#

1. Write the equa­tion in stan­dard form $ax^2 + bx + c = 0$.
2. Find two num­bers $m$ and $n$ with $m + n = b$ and $mn = ac$.
3. Rewrite $bx$ as $mx + nx$.
4. Group the first two terms and the last two terms, and take out the com­mon fac­tor from each group. The same bracket should appear twice.
5. Write the expres­sion as a prod­uct of two lin­ear fac­tors.
6. Set each fac­tor equal to zero and solve.
7. Check each root in the orig­i­nal equa­tion and, in a word prob­lem, reject any root that does not fit the sit­u­a­tion.

Worked exam­ples#

Exam­ple 1: two ratio­nal roots#

Find the roots of $2x^2 - 5x + 3 = 0$ by fac­tori­sa­tion.

Here $a = 2$, $b = -5$, $c = 3$, so $ac = 6$. We need two num­bers with sum $-5$ and prod­uct $6$: they are $-2$ and $-3$.

$$\begin{aligned} 2x^2 - 5x + 3 &= 2x^2 - 2x - 3x + 3 \\ &= 2x(x - 1) - 3(x - 1) \\ &= (2x - 3)(x - 1) \end{aligned}$$

So $(2x - 3)(x - 1) = 0$, which gives $2x - 3 = 0$ or $x - 1 = 0$.

Answer: the roots are $\displaystyle x = \frac{3}{2}$ and $x = 1$.

Check with $\displaystyle x = \frac{3}{2}$: $\displaystyle 2 \times \frac{9}{4} - 5 \times \frac{3}{2} + 3 = \frac{9}{2} - \frac{15}{2} + 3 = -3 + 3 = 0$.

Exam­ple 2: a neg­a­tive coef­fi­cient#

Find the roots of $6x^2 - x - 2 = 0$.

Here $ac = 6 \times (-2) = -12$ and $b = -1$. Two num­bers with sum $-1$ and prod­uct $-12$ are $3$ and $-4$.

$$\begin{aligned} 6x^2 - x - 2 &= 6x^2 + 3x - 4x - 2 \\ &= 3x(2x + 1) - 2(2x + 1) \\ &= (3x - 2)(2x + 1) \end{aligned}$$

So $3x - 2 = 0$ or $2x + 1 = 0$.

Answer: the roots are $\displaystyle x = \frac{2}{3}$ and $\displaystyle x = -\frac{1}{2}$.

Exam­ple 3: a repeated root with surds#

Find the roots of $3x^2 - 2\sqrt{6}\,x + 2 = 0$.

Here $ac = 3 \times 2 = 6$ and $b = -2\sqrt{6}$. Two num­bers with sum $-2\sqrt{6}$ and prod­uct $6$ are $-\sqrt{6}$ and $-\sqrt{6}$, since $(-\sqrt{6})(-\sqrt{6}) = 6$. Also note that $\sqrt{6} = \sqrt{3}\,\sqrt{2}$ and $3 = \sqrt{3}\,\sqrt{3}$.

$$\begin{aligned} 3x^2 - 2\sqrt{6}\,x + 2 &= 3x^2 - \sqrt{6}\,x - \sqrt{6}\,x + 2 \\ &= \sqrt{3}\,x\left(\sqrt{3}\,x - \sqrt{2}\right) - \sqrt{2}\left(\sqrt{3}\,x - \sqrt{2}\right) \\ &= \left(\sqrt{3}\,x - \sqrt{2}\right)\left(\sqrt{3}\,x - \sqrt{2}\right) \end{aligned}$$

Both fac­tors are the same, so the equa­tion is $\left(\sqrt{3}\,x - \sqrt{2}\right)^2 = 0$, and

$$\displaystyle \sqrt{3}\,x - \sqrt{2} = 0 \implies x = \frac{\sqrt{2}}{\sqrt{3}} = \frac{\sqrt{2} \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}} = \frac{\sqrt{6}}{3}$$

Answer: the roots are $\displaystyle \frac{\sqrt{6}}{3}$ and $\displaystyle \frac{\sqrt{6}}{3}$.

Because the same fac­tor appears twice, the root is counted twice. We say the equa­tion has two equal roots (a repeated root). On a graph, the curve touches the $x$-axis at one point with­out cross­ing it.

Exam­ple 4: the prayer hall#

In the pre­vi­ous les­son, the breadth $x$ m of a prayer hall of area 300 m², whose length is $(2x + 1)$ m, sat­is­fied $2x^2 + x - 300 = 0$. Find its dimen­sions.

Here $ac = 2 \times (-300) = -600$ and $b = 1$. Two num­bers with sum $1$ and prod­uct $-600$ are $25$ and $-24$.

$$\begin{aligned} 2x^2 + x - 300 &= 2x^2 - 24x + 25x - 300 \\ &= 2x(x - 12) + 25(x - 12) \\ &= (x - 12)(2x + 25) \end{aligned}$$

So $x = 12$ or $\displaystyle x = -\frac{25}{2} = -12.5$. The breadth is a length, so it can­not be neg­a­tive; we reject $-12.5$.

Answer: the breadth is 12 m and the length is $2(12) + 1 = 25$ m. Check: $12 \times 25 = 300$ m².

Exam­ple 5: the mar­bles#

John and Jivanti together had 45 mar­bles; after each lost 5, the prod­uct of their mar­bles was 124. With John's mar­bles as $x$, this gave $x^2 - 45x + 324 = 0$. How many did each have?

We need two num­bers with sum $-45$ and prod­uct $324$: they are $-9$ and $-36$.

$$\begin{aligned} x^2 - 45x + 324 &= x^2 - 9x - 36x + 324 \\ &= x(x - 9) - 36(x - 9) \\ &= (x - 9)(x - 36) \end{aligned}$$

So $x = 9$ or $x = 36$. Both are sen­si­ble here. If John had 9, Jivanti had $45 - 9 = 36$; if John had 36, Jivanti had 9.

Answer: one of them had 36 mar­bles and the other had 9. Check: after los­ing 5 each they have 4 and 31, and $4 \times 31 = 124$.

Com­mon mis­takes#

• Choos­ing two num­bers whose prod­uct is $c$ instead of $a \times c$ when $a \neq 1$.
• Drop­ping a sign while tak­ing out a com­mon fac­tor: from $-3x + 3$ the fac­tor is $-3$, giv­ing $-3(x - 1)$, not $-3(x + 1)$.
• Stop­ping at the fac­torised form and not writ­ing the roots.
• Apply­ing the zero-prod­uct rule when the right-hand side is not zero, for exam­ple writ­ing $x - 5 = 124$ from $(x - 5)(40 - x) = 124$.
• Divid­ing both sides by $x$ in an equa­tion such as $2x^2 = 5x$, which loses the root $x = 0$.
• Keep­ing a neg­a­tive length, age or num­ber of objects as the answer to a word prob­lem.

Try these#

1. Solve $x^2 - 3x - 10 = 0$. Answer: $x = 5$ or $x = -2$.
2. Solve $2x^2 + x - 6 = 0$. Answer: $\displaystyle x = \frac{3}{2}$ or $x = -2$.
3. Solve $\sqrt{2}\,x^2 + 7x + 5\sqrt{2} = 0$. Answer: $x = -\sqrt{2}$ or $\displaystyle x = -\frac{5\sqrt{2}}{2}$ (that is, $\displaystyle -\frac{5}{\sqrt{2}}$).
4. Solve $100x^2 - 20x + 1 = 0$. Answer: two equal roots, $\displaystyle x = \frac{1}{10}$ and $\displaystyle \frac{1}{10}$.
5. In the toy prob­lem, the num­ber of toys $x$ sat­is­fies $x^2 - 55x + 750 = 0$. Find $x$. Answer: $x = 25$ or $x = 30$.

Key terms#

Root (solu­tion)

A real num­ber $\alpha$ with $a\alpha^2 + b\alpha + c = 0$.

Zero of a poly­no­mial

A value of $x$ at which the poly­no­mial equals zero; the zeroes of $ax^2 + bx + c$ are the roots of $ax^2 + bx + c = 0$.

Fac­tori­sa­tion

Writ­ing an expres­sion as a prod­uct of sim­pler expres­sions, here two lin­ear fac­tors.

Split­ting the mid­dle term

Writ­ing $bx$ as $mx + nx$ with $m + n = b$ and $mn = ac$, so the expres­sion can be grouped.

Zero-prod­uct rule

If $pq = 0$ then $p = 0$ or $q = 0$.

Equal (repeated) roots

Two roots that are the same num­ber, aris­ing when the two fac­tors are iden­ti­cal.

Surd

An irra­tional root such as $\sqrt{2}$ or $\sqrt{6}$ left in exact form.

Com­mon ques­tions#

How many roots can a qua­dratic equa­tion have?#

At most two. It may have two dif­fer­ent real roots, two equal real roots, or no real roots at all.

What if I can­not find the two num­bers for the split?#

The equa­tion may not fac­torise neatly over ratio­nal num­bers, or it may have no real roots. In that case use the qua­dratic for­mula, and check the dis­crim­i­nant first.

Why must the equa­tion equal zero before fac­toris­ing?#

The zero-prod­uct rule only works for a prod­uct equal to zero. A prod­uct equal to 124 tells you noth­ing about each fac­tor sep­a­rately.

Do both roots always give an answer to a word prob­lem?#

No. Both sat­isfy the equa­tion, but a neg­a­tive breadth, age or count must be rejected. In the mar­bles prob­lem, how­ever, both roots give valid answers.

Is $\displaystyle \frac{\sqrt{2}}{\sqrt{3}}$ an accept­able final answer?#

It is cor­rect, but it is usual to ratio­nalise the denom­i­na­tor and write $\displaystyle \frac{\sqrt{6}}{3}$.

Ref­er­ences#

1. National Coun­cil of Edu­ca­tional Research and Train­ing. Math­e­mat­ics: Text­book for Class X. NCERT, New Delhi.
2. Gelfand, I. M. and Shen, A. Alge­bra. Birkhäuser.
3. Aggar­wal, R. S. Sec­ondary School Math­e­mat­ics for Class 10. Bharati Bhawan.