An Equation With a Squared Term Can Have Two Answers and You Must Reject One

Because squaring destroys information about sign. **Both $3$ and $-3$ square to $9$**, so an equation built on a square can be satisfied from two directions at once.

That single difference changes how you work. A linear equation such as $2x+6=0$ has exactly one answer, and once you find it you are finished. A quadratic equation gives you two candidates, and finishing means deciding what to do with both of them. Sometimes both are valid. Sometimes one of them is a length of $-7.5$ metres or an age of $-9$ years, and you must say so in writing and discard it.

The chapter is built in three stages, and this part covers the first two.

- Recognise whether an equation is genuinely quadratic, and write it in the standard form $a{x}^2+bx+c=0$ with $a\ne 0$
- Solve it by factorisation, splitting the middle term exactly as you did with polynomials
- Model a real situation with a quadratic, solve it, and reject any root the situation cannot accept

The recognition step matters more than it looks. An equation can carry an $x^2$ on both sides and turn out to be linear once you simplify, and an equation with brackets can hide a quadratic completely. Until the equation is in standard form you do not know what you are looking at, and reading off $a$, $b$ and $c$ from an unsimplified equation is the commonest early error in this chapter.

This page covers the first part of the CBSE Class 10 Maths chapter on quadratic equations: the standard form, solving by factorisation, and word problems where a root must be rejected.

**Expand every bracket, bring all the terms to one side, simplify, and then look at the highest power. It is quadratic only if an $x^2$ term survives with a non-zero coefficient.**

The standard form is

$$a{x}^2+bx+c=0,a\ne 0$$

where $a$ is the **coefficient of $x^2$**, $b$ the **coefficient of $x$** and $c$ the constant term. **The condition $a\ne 0$ is part of the definition**, not a footnote — with $a=0$ the equation is linear and has only one root.

Worked example 1. Is $(x-2{)}^2+1=2x-3$ a quadratic equation?

Expand the left side:

$$x^2-4x+4+1=2x-3$$
$$x^2-4x+5=2x-3$$

Bring everything to the left:

$$x^2-4x+5-2x+3=0⟹x^2-6x+8=0$$

Yes, it is quadratic, with $a=1$, $b=-6$ and $c=8$.

Worked example 2 — the equation that looks quadratic and is not. Is $x(x+1)+8=(x+2)(x-2)$ quadratic?

$$x^2+x+8=x^2-4$$

The $x^2$ terms are identical on both sides, so they cancel:

$$x+12=0$$

No — it is linear, with the single root $x=-12$. Check: $-12\times (-11)+8=132+8=140$, and $(-10)(-14)=140$. Both sides agree, so the linear answer is correct and there is no second root to look for.

That example is worth remembering. Two $x^2$ terms with the same coefficient on opposite sides always cancel, and an equation is judged after simplifying, never before.

Worked example 3. Is $x^2-2x=(-2)(3-x)$ quadratic?

$$x^2-2x=-6+2x⟹x^2-4x+6=0$$

Yes, with $a=1$, $b=-4$, $c=6$.

Worked example 4 — a fractional equation. Write $\frac{1}{x}+\frac{1}{x-3}=\frac{1}{2}$ in standard form, for $x\ne 0$ and $x\ne 3$.

Multiplying through by $2x(x-3)$:

$$2(x-3)+2x=x(x-3)$$
$$2x-6+2x=x^2-3x$$
$$0=x^2-7x+6$$

So it is quadratic in disguise, with $a=1$, $b=-7$, $c=6$.

Notice the restriction written alongside it. The original equation is meaningless at $x=0$ and $x=3$, so those values would have to be discarded even if they appeared as roots. They do not here — the roots of $x^2-7x+6=0$ are $1$ and $6$ — but stating the restriction is part of a complete answer.

Signs are where marks disappear. In $x^2-6x+8=0$ the values are $b=-6$ and $c=8$; writing $b=6$ because the number printed is $6$ is the error that then ruins every later calculation. Read the sign that sits in front of the term.

**Find two numbers whose product is $a\times c$ and whose sum is $b$, split the middle term into those two pieces, group in pairs, and take out the common factor. Then set each factor to zero.

Why setting each factor to zero is valid. If a product of two quantities is zero, at least one of them must be zero. That is the entire logic of the method**, and it is worth stating in an answer: from $(x-5)(x+2)=0$ it follows that $x-5=0$ or $x+2=0$.

Worked example 1. Solve $2x^2-5x+3=0$.

Here $a\times c=2\times 3=6$ and $b=-5$, so we need two numbers with product $6$ and sum $-5$: those are $-2$ and $-3$.

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

So $x=1$ or $x=\frac{3}{2}$.

Check both. At $x=1$: $2-5+3=0$. At $x=\frac{3}{2}$: $2\times \frac{9}{4}-5\times \frac{3}{2}+3=\frac{9}{2}-\frac{15}{2}+3=-3+3=0$. Both roots satisfy the equation.

Worked example 2. Solve $6x^2-x-2=0$.

Now $a\times c=6\times (-2)=-12$ and $b=-1$, so we need product $-12$ and sum $-1$: those are $-4$ and $3$.

$$6x^2-4x+3x-2=0$$
$$2x(3x-2)+1(3x-2)=0$$
$$(3x-2)(2x+1)=0$$

So $x=\frac{2}{3}$ or $x=-\frac{1}{2}$.

**Check at $x=\frac{2}{3}$**: $6\times \frac{4}{9}-\frac{2}{3}-2=\frac{8}{3}-\frac{2}{3}-2=2-2=0$. Correct.

Worked example 3 — equal roots. Solve $100x^2-20x+1=0$.

Product $=100$ and sum $=-20$, so both numbers are $-10$:

$$100x^2-10x-10x+1=0$$
$$10x(10x-1)-1(10x-1)=0$$
$$(10x-1)(10x-1)=0⟹(10x-1{)}^2=0$$

**The only root is $x=\frac{1}{10}$, and it is called a repeated or equal root because the same factor appears twice.

Check**: $100\times \frac{1}{100}-20\times \frac{1}{10}+1=1-2+1=0$. Correct.

Worked example 4 — a common factor first. Solve $2x^2+x-6=0$.

Product $=-12$, sum $=1$, so the numbers are $4$ and $-3$:

$$2x^2+4x-3x-6=0$$
$$2x(x+2)-3(x+2)=0$$
$$(x+2)(2x-3)=0$$

So $x=-2$ or $x=\frac{3}{2}$. **Check at $x=-2$**: $8-2-6=0$. Correct.

The habit that makes splitting reliable. Write down $a\times c$ and $b$ as a pair, with their signs, before guessing anything. Then the two numbers are determined: if the product is negative the two numbers have opposite signs, and if the product is positive they share the sign of $b$. That rule turns guessing into a short search, and it is what separates a student who factorises in thirty seconds from one who tries pairs at random.

A root always means a factor. Once you have the roots you can rebuild the equation, which is the fastest possible check: roots $1$ and $\frac{3}{2}$ come from $(x-1)(2x-3)=2x^2-5x+3$, which is where we started. Rebuilding takes one line and catches every sign slip.

Name the unknown, write the condition as an equation, solve it, and then test each root against the meaning of the unknown. A length, an age, a count or a speed cannot be negative, and a count must be a whole number.

Worked example 1 — consecutive integers. Find two consecutive positive integers whose product is $306$.

Let the smaller integer be $x$, so the next one is $x+1$:

$$x(x+1)=306⟹x^2+x-306=0$$

Product $=-306$ and sum $=1$, so the numbers are $18$ and $-17$:

$$x^2+18x-17x-306=0$$
$$x(x+18)-17(x+18)=0$$
$$(x+18)(x-17)=0$$

So $x=17$ or $x=-18$.

Now the rejection, and it must be written out. The problem asks for positive integers, so $x=-18$ is inadmissible. **The integers are $17$ and $18$.

Check**: $17\times 18=306$. Correct.

Worked example 2 — an area. The length of a rectangular park is $3$ m more than twice its breadth, and its area is $90$ m${}^2$. Find its dimensions.

Let the breadth be $b$ metres, so the length is $2b+3$ metres:

$$b(2b+3)=90⟹2b^2+3b-90=0$$

Product $=2\times (-90)=-180$ and sum $=3$, so the numbers are $15$ and $-12$:

$$2b^2+15b-12b-90=0$$
$$b(2b+15)-6(2b+15)=0$$
$$(2b+15)(b-6)=0$$

So $b=6$ or $b=-\frac{15}{2}$.

**A breadth cannot be $-7.5$ m, so that root is rejected. The breadth is $6$ m and the length is $2(6)+3=15$ m.

Check**: $6\times 15=90$ m${}^2$, and $15$ is indeed $3$ more than twice $6$. Both conditions hold.

Worked example 3 — an age. The product of a boy's age two years ago and his age four years from now is $55$. Find his present age.

Let his present age be $x$ years:

$$(x-2)(x+4)=55⟹x^2+2x-8=55⟹x^2+2x-63=0$$

Product $=-63$ and sum $=2$, giving $9$ and $-7$:

$$(x+9)(x-7)=0$$

So $x=7$ or $x=-9$. An age cannot be negative, so $x=7$.

Check: two years ago he was $5$ and four years from now he will be $11$, and $5\times 11=55$. Correct.

Worked example 4 — two roots both valid. Two friends together have $45$ marbles. Each loses $5$ marbles, and the product of the numbers they now have is $124$. Find how many each had.

Let one friend have $x$ marbles, so the other has $45-x$:

$$(x-5)(45-x-5)=124⟹(x-5)(40-x)=124$$
$$40x-x^2-200+5x=124⟹x^2-45x+324=0$$

Product $=324$ and sum $=-45$, giving $-36$ and $-9$:

$$(x-36)(x-9)=0$$

So $x=36$ or $x=9$.

Here both roots are admissible, and they describe the same situation from the two friends' points of view: one had $36$ and the other $9$. Neither is rejected.

Check: $36+9=45$, and after losing $5$ each they hold $31$ and $4$, whose product is $124$. Correct.

So rejection is not automatic. The test is always the same question: does this value make sense for the quantity the letter stands for? A count of marbles of $9$ makes sense; a breadth of $-7.5$ m does not; and when both roots survive the test, both belong in the answer.

Worked example 5 — two numbers differing by a fixed amount. Two numbers differ by $3$ and their product is $504$. Find them.

Let the smaller be $x$, so the larger is $x+3$:

$$x(x+3)=504⟹x^2+3x-504=0$$

Product $=-504$ and sum $=3$, giving $24$ and $-21$:

$$(x+24)(x-21)=0$$

So $x=21$ or $x=-24$. If the question says positive numbers, the pair is $21$ and $24$; if it does not, the pair $-24$ and $-21$ is equally valid, since their difference is $3$ and their product is $504$. Read the wording before rejecting anything.