NCERT Class 10 Mathematics — Quadratic Equations, Exercise 4.1. Both questions solved.

Exercise 4.1 does not ask you to solve anything. It tests two skills that come before solving: recognising a quadratic equation when it is disguised, and building one from a situation described in words. Both rest on one definition.

A quadratic equation in $x$ is one that can be written in the standard form

$$ax^2 + bx + c = 0, \qquad a, b, c \text{ real},\ a \ne 0.$$

Only $a$ has to be non-zero. Either $b$ or $c$ (or both) may be $0$.

Key insight. Whether an equation is quadratic is decided by its simplified form, not by how it looks. Expand both sides, bring everything to one side and collect terms; then look at the highest power of $x$ that survives. An $x^2$ on each side can cancel and leave a linear equation, and an $x^3$ on each side can cancel and leave a quadratic.

Question 1

Check whether the following are quadratic equations:

(i) $(x + 1)^2 = 2(x – 3)$    (ii) $x^2 – 2x = (-2)(3 – x)$

(iii) $(x – 2)(x + 1) = (x – 1)(x + 3)$    (iv) $(x – 3)(2x + 1) = x(x + 5)$

(v) $(2x – 1)(x – 3) = (x + 5)(x – 1)$    (vi) $x^2 + 3x + 1 = (x – 2)^2$

(vii) $(x + 2)^3 = 2x(x^2 – 1)$    (viii) $x^3 – 4x^2 – x + 1 = (x – 2)^3$

Solution. In every part, expand, move all terms to the left, and read off the degree of what remains.

(i) The left side is $x^2 + 2x + 1$ and the right side is $2x – 6$, so

$$x^2 + 2x + 1 = 2x – 6 \quad\Longrightarrow\quad x^2 + 7 = 0$$

This is $ax^2 + bx + c = 0$ with $a = 1$, $b = 0$, $c = 7$. A missing $x$ term does not matter, because only $a$ must be non-zero. Quadratic.

(ii) The right side is $(-2)(3 – x) = -6 + 2x$, so

$$x^2 – 2x = -6 + 2x \quad\Longrightarrow\quad x^2 – 4x + 6 = 0$$

Quadratic.

(iii) Both sides begin with $x^2$:

$$x^2 – x – 2 = x^2 + 2x – 3 \quad\Longrightarrow\quad -3x + 1 = 0$$

The $x^2$ terms cancel, leaving a linear equation. Not quadratic.

(iv) The left side is $2x^2 + x – 6x – 3 = 2x^2 – 5x – 3$ and the right side is $x^2 + 5x$:

$$2x^2 – 5x – 3 = x^2 + 5x \quad\Longrightarrow\quad x^2 – 10x – 3 = 0$$

The $x^2$ terms do not cancel, because their coefficients $2$ and $1$ differ. Quadratic.

(v) The left side is $2x^2 – 7x + 3$ and the right side is $x^2 + 4x – 5$:

$$2x^2 – 7x + 3 = x^2 + 4x – 5 \quad\Longrightarrow\quad x^2 – 11x + 8 = 0$$

Quadratic.

(vi) The right side is $(x – 2)^2 = x^2 – 4x + 4$:

$$x^2 + 3x + 1 = x^2 – 4x + 4 \quad\Longrightarrow\quad 7x – 3 = 0$$

Again the $x^2$ terms cancel. Not quadratic.

(vii) The left side is $(x + 2)^3 = x^3 + 6x^2 + 12x + 8$ and the right side is $2x^3 – 2x$:

$$x^3 + 6x^2 + 12x + 8 = 2x^3 – 2x \quad\Longrightarrow\quad x^3 – 6x^2 – 14x – 8 = 0$$

The cubes do not cancel, since one side has $x^3$ and the other $2x^3$. The equation has degree $3$. Not quadratic.

(viii) The right side is $(x – 2)^3 = x^3 – 6x^2 + 12x – 8$:

$$x^3 – 4x^2 – x + 1 = x^3 – 6x^2 + 12x – 8 \quad\Longrightarrow\quad 2x^2 – 13x + 9 = 0$$

This time both sides carry exactly $x^3$, so the cubes cancel and a genuine quadratic is left behind. Quadratic.

Question 2

Represent the following situations in the form of quadratic equations:

(i) The area of a rectangular plot is $528\ \text{m}^2$. The length of the plot (in metres) is one more than twice its breadth. We need to find the length and breadth of the plot.

(ii) The product of two consecutive positive integers is $306$. We need to find the integers.

(iii) Rohan’s mother is $26$ years older than him. The product of their ages (in years) $3$ years from now will be $360$. We would like to find Rohan’s present age.

(iv) A train travels a distance of $480$ km at a uniform speed. If the speed had been $8$ km/h less, then it would have taken $3$ hours more to cover the same distance. We need to find the speed of the train.

Solution. In each part, choose one unknown, write every other quantity in terms of it, and turn the one fact that links them into an equation.

(i) Let the breadth be $x$ m. “One more than twice the breadth” makes the length $(2x + 1)$ m. Area is length times breadth:

$$x(2x + 1) = 528 \quad\Longrightarrow\quad 2x^2 + x – 528 = 0$$

(ii) Consecutive integers differ by $1$, so let them be $x$ and $x + 1$:

$$x(x + 1) = 306 \quad\Longrightarrow\quad x^2 + x – 306 = 0$$

(iii) Let Rohan’s present age be $x$ years, so his mother is $x + 26$. In $3$ years both will be $3$ years older: $x + 3$ and $x + 29$. Hence

$$(x + 3)(x + 29) = 360 \quad\Longrightarrow\quad x^2 + 32x + 87 = 360 \quad\Longrightarrow\quad x^2 + 32x – 273 = 0$$

(iv) Let the speed be $u$ km/h. Time is distance over speed, so the journey takes $\dfrac{480}{u}$ hours, and at the slower speed $\dfrac{480}{u – 8}$ hours. The slower journey takes $3$ hours longer:

$$\frac{480}{u – 8} – \frac{480}{u} = 3$$

Multiplying through by $u(u – 8)$ (neither factor is zero, since the train moves at more than $8$ km/h):

$$480u – 480(u – 8) = 3u(u – 8) \quad\Longrightarrow\quad 3840 = 3u^2 – 24u \quad\Longrightarrow\quad u^2 – 8u – 1280 = 0$$

As a check that each equation really captures its situation: solved, they give a breadth of $16$ m (so the plot is $16 \times 33 = 528\ \text{m}^2$), the integers $17$ and $18$, a present age of $7$ years for Rohan ($10 \times 36 = 360$), and a speed of $40$ km/h ($12$ hours against $15$ hours at $32$ km/h).

Common mistakes

• Question 1(i), rejecting $x^2 + 7 = 0$. Some students reject it because there is no $x$ term, others because it has no real solution. Neither matters. The definition asks only that the coefficient of $x^2$ be non-zero, and it is $1$.
• Questions 1(iii) and 1(vi), deciding at a glance. Both look quadratic because $x^2$ appears on each side. After expanding, the $x^2$ terms cancel and what is left is linear. You cannot judge without simplifying first.
• Questions 1(vii) and 1(viii), treating them alike. Both contain cubes on both sides, but only in (viii) are the $x^3$ coefficients equal, so only there do they cancel. In (vii) the $x^3$ on the left and the $2x^3$ on the right leave $-x^3$ behind.
• Question 2(i), writing the length as $2(x + 1)$. “One more than twice its breadth” means double first, then add one: $2x + 1$. The bracketed version is “twice one more than the breadth”, which is a different plot.
• Question 2(iii), adding $3$ years to only one age. “Three years from now” applies to Rohan and his mother alike. Writing $(x + 3)(x + 26) = 360$ ages Rohan but freezes his mother.
• Question 2(iv), subtracting the times the wrong way round. The slower train takes longer, so it is $\frac{480}{u – 8}$ minus $\frac{480}{u}$ that equals $3$. Reversing them gives $u^2 – 8u + 1280 = 0$, which has no real roots at all. That is a sign something went wrong in the set-up.

Practise next

• Exercise 4.2 — factorisation, which solves the four equations built in question 2 and turns them back into lengths, integers, ages and speeds.
• Chapter 2, Exercise 2.2 — zeroes of a quadratic polynomial, which are exactly the roots of the corresponding quadratic equation, seen from the polynomial side.