NCERT Class 10 Exercise 4.1 Solutions: Quadratic Equations

Ques­tion 1#

Check whether the fol­low­ing are qua­dratic equa­tions.

Ques­tion 1 (i)#

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

Solu­tion. Expand the left-hand side with $(a+b)^2 = a^2 + 2ab + b^2$ and the right-hand side by dis­trib­ut­ing:

$$x^2 + 2x + 1 = 2x - 6$$

Bring every term to the left:

$$x^2 + 2x + 1 - 2x + 6 = 0 \implies x^2 + 7 = 0$$

The $2x$ terms can­cel, but $x^2$ remains with coef­fi­cient 1. This is $ax^2 + bx + c = 0$ with $a = 1$, $b = 0$, $c = 7$.

Answer: Yes, it is a qua­dratic equa­tion ($x^2 + 7 = 0$).

Ques­tion 1 (ii)#

$x^2 - 2x = (-2)(3 - x)$

Solu­tion. Expand the right-hand side: $(-2)(3 - x) = -6 + 2x$. So

$$x^2 - 2x = -6 + 2x$$

Bring every term to the left:

$$x^2 - 2x - 2x + 6 = 0 \implies x^2 - 4x + 6 = 0$$

An $x^2$ term with coef­fi­cient $1 \neq 0$ sur­vives.

Answer: Yes, it is a qua­dratic equa­tion ($x^2 - 4x + 6 = 0$).

Ques­tion 1 (iii)#

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

Solu­tion. Both sides con­tain $x^2$, so expand both before decid­ing.

$$\begin{aligned} \text{LHS} &= x^2 + x - 2x - 2 = x^2 - x - 2 \\ \text{RHS} &= x^2 + 3x - x - 3 = x^2 + 2x - 3 \end{aligned}$$

Sub­tract the right-hand side from the left-hand side:

$$x^2 - x - 2 - x^2 - 2x + 3 = 0 \implies -3x + 1 = 0$$

The $x^2$ terms can­cel exactly and no $x^2$ term is left. The equa­tion is lin­ear.

Answer: No, it is not a qua­dratic equa­tion (it reduces to the lin­ear equa­tion $-3x + 1 = 0$).

Ques­tion 1 (iv)#

$(x - 3)(2x + 1) = x(x + 5)$

Solu­tion.

$$\begin{aligned} \text{LHS} &= 2x^2 + x - 6x - 3 = 2x^2 - 5x - 3 \\ \text{RHS} &= x^2 + 5x \end{aligned}$$

Sub­tract­ing:

$$2x^2 - 5x - 3 - x^2 - 5x = 0 \implies x^2 - 10x - 3 = 0$$

The squared terms were $2x^2$ and $x^2$, so they do not can­cel; $x^2$ sur­vives with coef­fi­cient 1.

Answer: Yes, it is a qua­dratic equa­tion ($x^2 - 10x - 3 = 0$).

Ques­tion 1 (v)#

$(2x - 1)(x - 3) = (x + 5)(x - 1)$

Solu­tion.

$$\begin{aligned} \text{LHS} &= 2x^2 - 6x - x + 3 = 2x^2 - 7x + 3 \\ \text{RHS} &= x^2 - x + 5x - 5 = x^2 + 4x - 5 \end{aligned}$$

Sub­tract­ing:

$$2x^2 - 7x + 3 - x^2 - 4x + 5 = 0 \implies x^2 - 11x + 8 = 0$$

Again $2x^2 - x^2 = x^2 \neq 0$.

Answer: Yes, it is a qua­dratic equa­tion ($x^2 - 11x + 8 = 0$).

Ques­tion 1 (vi)#

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

Solu­tion. Expand the right-hand side: $(x - 2)^2 = x^2 - 4x + 4$. So

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

Sub­tract­ing:

$$x^2 + 3x + 1 - x^2 + 4x - 4 = 0 \implies 7x - 3 = 0$$

Both sides had exactly $x^2$, so the squared terms can­cel and only a lin­ear equa­tion remains.

Answer: No, it is not a qua­dratic equa­tion (it reduces to $7x - 3 = 0$).

Ques­tion 1 (vii)#

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

Solu­tion. Use $(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$ with $a = x$, $b = 2$:

$$(x+2)^3 = x^3 + 3x^2(2) + 3x(4) + 8 = x^3 + 6x^2 + 12x + 8$$

The right-hand side is $2x(x^2 - 1) = 2x^3 - 2x$. Sub­tract­ing:

$$x^3 + 6x^2 + 12x + 8 - 2x^3 + 2x = 0 \implies -x^3 + 6x^2 + 14x + 8 = 0$$

Mul­ti­ply­ing by $-1$: $x^3 - 6x^2 - 14x - 8 = 0$. The $x^3$ terms were $x^3$ and $2x^3$, so they do not can­cel; the equa­tion has degree 3.

Answer: No, it is not a qua­dratic equa­tion (it is the cubic equa­tion $x^3 - 6x^2 - 14x - 8 = 0$).

Ques­tion 1 (viii)#

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

Solu­tion. Use $(a-b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$ with $a = x$, $b = 2$:

$$(x-2)^3 = x^3 - 6x^2 + 12x - 8$$

So the equa­tion is $x^3 - 4x^2 - x + 1 = x^3 - 6x^2 + 12x - 8$. Sub­tract­ing:

$$\begin{aligned} x^3 - 4x^2 - x + 1 - x^3 + 6x^2 - 12x + 8 &= 0 \\ 2x^2 - 13x + 9 &= 0 \end{aligned}$$

Both sides started with $x^3$ with the same coef­fi­cient 1, so the cubes can­cel, and $x^2$ sur­vives with coef­fi­cient $2 \neq 0$.

Answer: Yes, it is a qua­dratic equa­tion ($2x^2 - 13x + 9 = 0$).

Why (vii) and (viii) go dif­fer­ent ways. They look alike, but in (vii) the cubic terms are $x^3$ and $2x^3$, which leave $-x^3$ behind, while in (viii) both sides have exactly $x^3$, which can­cel and drop the degree to 2.

Sum­mary for Ques­tion 1: qua­dratic: (i), (ii), (iv), (v), (viii). Not qua­dratic: (iii) and (vi) are lin­ear, and (vii) is cubic.

Ques­tion 2#

Rep­re­sent the fol­low­ing sit­u­a­tions in the form of qua­dratic equa­tions.

The ques­tion asks only for the equa­tion. After each one, we also solve it and check the answer against the story, so that you can be sure the equa­tion is right.

Ques­tion 2 (i)#

The area of a rec­tan­gu­lar plot is 528 m². 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.

Solu­tion. Let the breadth of the plot be $x$ m. Then the length is $(2x + 1)$ m. Since area = length × breadth,

$$\begin{aligned} x(2x + 1) &= 528 \\ 2x^2 + x - 528 &= 0 \end{aligned}$$

Answer: $2x^2 + x - 528 = 0$, where $x$ m is the breadth.

Going fur­ther. We need two num­bers with prod­uct $2 \times (-528) = -1056$ and sum $1$. These are $33$ and $-32$. Split­ting the mid­dle term:

$$\begin{aligned} 2x^2 + 33x - 32x - 528 &= 0 \\ x(2x + 33) - 16(2x + 33) &= 0 \\ (2x + 33)(x - 16) &= 0 \end{aligned}$$

So $x = 16$ or $\displaystyle x = -\frac{33}{2}$. A breadth can­not be neg­a­tive, so the breadth is 16 m and the length is $2(16) + 1 = 33$ m. Check: $16 \times 33 = 528$ m², and 33 is one more than twice 16.

Ques­tion 2 (ii)#

The prod­uct of two con­sec­u­tive pos­i­tive inte­gers is 306. We need to find the inte­gers.

Solu­tion. Let the smaller inte­ger be $x$. The next con­sec­u­tive inte­ger is $x + 1$. Their prod­uct is 306:

$$\begin{aligned} x(x + 1) &= 306 \\ x^2 + x - 306 &= 0 \end{aligned}$$

Answer: $x^2 + x - 306 = 0$, where $x$ is the smaller inte­ger.

Going fur­ther. Two num­bers with prod­uct $-306$ and sum $1$ are $18$ and $-17$.

$$\begin{aligned} x^2 + 18x - 17x - 306 &= 0 \\ x(x + 18) - 17(x + 18) &= 0 \\ (x + 18)(x - 17) &= 0 \end{aligned}$$

So $x = 17$ or $x = -18$. The inte­gers are pos­i­tive, so $x = 17$ and the inte­gers are 17 and 18. Check: $17 \times 18 = 306$.

Ques­tion 2 (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.

Solu­tion. Let Rohan's present age be $x$ years. His moth­er's present age is $(x + 26)$ years. After 3 years, Rohan will be $(x + 3)$ years old and his mother $(x + 29)$ years old. Their prod­uct then is 360:

$$\begin{aligned} (x + 3)(x + 29) &= 360 \\ x^2 + 29x + 3x + 87 &= 360 \\ x^2 + 32x - 273 &= 0 \end{aligned}$$

Answer: $x^2 + 32x - 273 = 0$, where $x$ years is Rohan's present age.

Going fur­ther. Two num­bers with prod­uct $-273$ and sum $32$ are $39$ and $-7$.

$$\begin{aligned} x^2 + 39x - 7x - 273 &= 0 \\ x(x + 39) - 7(x + 39) &= 0 \\ (x + 39)(x - 7) &= 0 \end{aligned}$$

So $x = 7$ or $x = -39$. An age can­not be neg­a­tive, so Rohan is 7 years old and his mother is 33. Check: in 3 years they will be 10 and 36, and $10 \times 36 = 360$.

Ques­tion 2 (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. Let the speed of the train be $x$ km/h. Since time = dis­tance ÷ speed, the time taken is $\displaystyle \frac{480}{x}$ hours. At $(x - 8)$ km/h the time would be $\displaystyle \frac{480}{x - 8}$ hours, which is 3 hours more:

$$\displaystyle \frac{480}{x - 8} - \frac{480}{x} = 3$$

Com­bine the frac­tions over the com­mon denom­i­na­tor $x(x - 8)$:

$$\displaystyle \begin{aligned} \frac{480x - 480(x - 8)}{x(x - 8)} &= 3 \\ \frac{3840}{x^2 - 8x} &= 3 \\ 3840 &= 3x^2 - 24x \\ 3x^2 - 24x - 3840 &= 0 \end{aligned}$$

Divid­ing every term by 3:

$$x^2 - 8x - 1280 = 0$$

Answer: $x^2 - 8x - 1280 = 0$, where $x$ km/h is the speed of the train.

Going fur­ther. Two num­bers with prod­uct $-1280$ and sum $-8$ are $-40$ and $32$.

$$\begin{aligned} x^2 - 40x + 32x - 1280 &= 0 \\ x(x - 40) + 32(x - 40) &= 0 \\ (x - 40)(x + 32) &= 0 \end{aligned}$$

So $x = 40$ or $x = -32$. A speed can­not be neg­a­tive, so the speed is 40 km/h. Check: $\displaystyle \frac{480}{40} = 12$ h and $\displaystyle \frac{480}{32} = 15$ h, a dif­fer­ence of 3 h.

Why this works: mul­ti­ply­ing both sides by $x(x - 8)$ is allowed because nei­ther $x$ nor $x - 8$ can be zero for a real train speed, and it turns the frac­tional equa­tion into a qua­dratic one.