NCERT Class 10 Exercise 4.3 Solutions: Nature of Roots

These are com­plete solu­tions to NCERT Class 10 Math­e­mat­ics, Exer­cise 4.3 (Chap­ter 4, Qua­dratic Equa­tions, Sec­tion 4.4 Nature of Roots). The exer­cise prac­tises using the dis­crim­i­nant: first to describe the roots of given equa­tions, then to find an unknown coef­fi­cient, and finally to decide whether three real-life sit­u­a­tions are pos­si­ble at all.

Quick recap of the method#

For $ax^2 + bx + c = 0$ with $a \neq 0$, the dis­crim­i­nant is $D = b^2 - 4ac$, and

• $D \gt 0$: two dis­tinct real roots, $\displaystyle x = \frac{-b \pm \sqrt{D}}{2a}$;
• $D = 0$: two equal real roots, each $\displaystyle x = -\frac{b}{2a}$;
• $D \lt 0$: no real roots.

1. Write the equa­tion in stan­dard form and read off $a$, $b$, $c$ with their signs.
2. Com­pute $D$ and decide the nature of the roots.
3. If the roots are real, find them.
4. In a word prob­lem, a neg­a­tive $D$ means the sit­u­a­tion is not pos­si­ble; oth­er­wise, keep only roots that make sense.

Ques­tion 1#

Find the nature of the roots of the fol­low­ing qua­dratic equa­tions. If the real roots exist, find them.

Ques­tion 1 (i)#

$2x^2 - 3x + 5 = 0$

Solu­tion. Here $a = 2$, $b = -3$, $c = 5$.

$$D = b^2 - 4ac = (-3)^2 - 4(2)(5) = 9 - 40 = -31$$

Since $D \lt 0$, the equa­tion has no real roots.

Why this works: com­plet­ing the square gives $\displaystyle 2x^2 - 3x + 5 = 2\left(x - \frac{3}{4}\right)^2 + \frac{31}{8}$, which is at least $\displaystyle \frac{31}{8}$ for every real $x$, so it can never be 0.

Answer: no real roots.

Ques­tion 1 (ii)#

$3x^2 - 4\sqrt{3}\,x + 4 = 0$

Solu­tion. Here $a = 3$, $b = -4\sqrt{3}$, $c = 4$.

$$D = \left(-4\sqrt{3}\right)^2 - 4(3)(4) = 16 \times 3 - 48 = 48 - 48 = 0$$

Since $D = 0$, the roots are real and equal, each being

$$\displaystyle x = -\frac{b}{2a} = \frac{4\sqrt{3}}{2 \times 3} = \frac{2\sqrt{3}}{3} = \frac{2}{\sqrt{3}}$$

Check: with $\displaystyle x = \frac{2\sqrt{3}}{3}$, $\displaystyle x^2 = \frac{12}{9} = \frac{4}{3}$, so $3x^2 = 4$; and $\displaystyle 4\sqrt{3}\,x = \frac{8 \times 3}{3} = 8$. Then $4 - 8 + 4 = 0$.

Answer: two equal real roots, $\displaystyle \frac{2}{\sqrt{3}}$ and $\displaystyle \frac{2}{\sqrt{3}}$ (that is, $\displaystyle \frac{2\sqrt{3}}{3}$ each).

Ques­tion 1 (iii)#

$2x^2 - 6x + 3 = 0$

Solu­tion. Here $a = 2$, $b = -6$, $c = 3$.

$$D = (-6)^2 - 4(2)(3) = 36 - 24 = 12$$

Since $D \gt 0$, there are two dis­tinct real roots. As 12 is not a per­fect square, they are irra­tional. Using $\sqrt{12} = 2\sqrt{3}$:

$$\displaystyle x = \frac{-b \pm \sqrt{D}}{2a} = \frac{6 \pm 2\sqrt{3}}{4} = \frac{3 \pm \sqrt{3}}{2}$$

Check: the sum of the roots is $\displaystyle \frac{3 + \sqrt{3}}{2} + \frac{3 - \sqrt{3}}{2} = 3 = -\frac{b}{a}$, and the prod­uct is $\displaystyle \frac{(3 + \sqrt{3})(3 - \sqrt{3})}{4} = \frac{9 - 3}{4} = \frac{3}{2} = \frac{c}{a}$.

Answer: two dis­tinct real roots, $\displaystyle x = \frac{3 + \sqrt{3}}{2}$ and $\displaystyle x = \frac{3 - \sqrt{3}}{2}$ (about 2.37 and 0.63, tak­ing $\sqrt{3} \approx 1.732$).

Ques­tions 3 to 5: is the sit­u­a­tion pos­si­ble?#

In each of these, form a qua­dratic equa­tion, use the dis­crim­i­nant to decide whether real roots exist, and then find the answer if they do.

Ques­tion 3#

Is it pos­si­ble to design a rec­tan­gu­lar mango grove whose length is twice its breadth, and the area is 800 m²? If so, find its length and breadth.

Solu­tion. Let the breadth be $x$ m. Then the length is $2x$ m, and the area con­di­tion gives

$$2x \times x = 800 \implies 2x^2 - 800 = 0$$

Here $a = 2$, $b = 0$, $c = -800$, so

$$D = 0^2 - 4(2)(-800) = 6400 \gt 0$$

Real roots exist, so the design is pos­si­ble. To find them, $2x^2 = 800$ gives $x^2 = 400$, so $x = 20$ or $x = -20$. A breadth can­not be neg­a­tive, so $x = 20$.

Check: $40 \times 20 = 800$ m², and 40 is twice 20.

Answer: yes, it is pos­si­ble. Breadth = 20 m, length = 40 m.

Ques­tion 4#

Is the fol­low­ing sit­u­a­tion pos­si­ble? If so, deter­mine their present ages. The sum of the ages of two friends is 20 years. Four years ago, the prod­uct of their ages in years was 48.

Solu­tion. Let the present age of one friend be $x$ years. Then the other is $(20 - x)$ years old. Four years ago their ages were $(x - 4)$ and $(16 - x)$ years, so

$$\begin{aligned} (x - 4)(16 - x) &= 48 \\ 16x - x^2 - 64 + 4x &= 48 \\ -x^2 + 20x - 112 &= 0 \\ x^2 - 20x + 112 &= 0 \end{aligned}$$

Here $a = 1$, $b = -20$, $c = 112$:

$$D = (-20)^2 - 4(1)(112) = 400 - 448 = -48$$

Since $D \lt 0$, the equa­tion has no real roots, so no real ages sat­isfy both con­di­tions.

Why this works: four years ago the two ages added up to $(x - 4) + (16 - x) = 12$. Two num­bers with sum 12 have the largest prod­uct when both are 6, and $6 \times 6 = 36$. Since 48 is more than 36, the prod­uct 48 can never be reached.

Answer: no, the sit­u­a­tion is not pos­si­ble.

Ques­tion 5#

Is it pos­si­ble to design a rec­tan­gu­lar park of perime­ter 80 m and area 400 m²? If so, find its length and breadth.

Solu­tion. Let the length be $x$ m. Since the perime­ter is $2(\text{length} + \text{breadth}) = 80$, length + breadth = 40, so the breadth is $(40 - x)$ m. The area con­di­tion gives

$$\begin{aligned} x(40 - x) &= 400 \\ 40x - x^2 &= 400 \\ x^2 - 40x + 400 &= 0 \end{aligned}$$

Here $a = 1$, $b = -40$, $c = 400$:

$$D = (-40)^2 - 4(1)(400) = 1600 - 1600 = 0$$

Since $D = 0$, the equa­tion has two equal real roots, so the park is pos­si­ble. The root is

$$\displaystyle x = -\frac{b}{2a} = \frac{40}{2} = 20$$

So the length is 20 m and the breadth is $40 - 20 = 20$ m. The park is in fact a square.

Check: perime­ter $= 2(20 + 20) = 80$ m and area $= 20 \times 20 = 400$ m².

Answer: yes, it is pos­si­ble. Length = 20 m and breadth = 20 m. Because $D = 0$, this is the only rec­tan­gle that meets both con­di­tions.

Key terms#

Dis­crim­i­nant

$D = b^2 - 4ac$ for the equa­tion $ax^2 + bx + c = 0$.

Nature of roots

Whether the roots are real and dis­tinct ($D \gt 0$), real and equal ($D = 0$), or not real ($D \lt 0$).

Equal roots

Two iden­ti­cal real roots, each equal to $\displaystyle -\frac{b}{2a}$.

Irra­tional roots

Real roots that involve a surd, as when $D$ is pos­i­tive but not a per­fect square.

Qua­dratic for­mula

$\displaystyle x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, valid when $b^2 - 4ac \geq 0$.

Sum and prod­uct of roots

For $ax^2 + bx + c = 0$, the roots add to $\displaystyle -\frac{b}{a}$ and mul­ti­ply to $\displaystyle \frac{c}{a}$; a quick check on answers.

Perime­ter

The total bound­ary length of a rec­tan­gle, $2(\text{length} + \text{breadth})$.

Com­mon ques­tions#

Why is $k = 0$ rejected in Ques­tion 2 (ii)?#

With $k = 0$ the $x^2$ term van­ishes and the equa­tion becomes $6 = 0$, which is not a qua­dratic equa­tion at all. The con­di­tion for equal roots applies only to qua­drat­ics.

Is $\displaystyle \frac{2}{\sqrt{3}}$ the same as $\displaystyle \frac{2\sqrt{3}}{3}$?#

Yes. Mul­ti­ply­ing the numer­a­tor and denom­i­na­tor of $\displaystyle \frac{2}{\sqrt{3}}$ by $\sqrt{3}$ gives $\displaystyle \frac{2\sqrt{3}}{3}$. Either form is accepted.

In Ques­tion 3, why use the dis­crim­i­nant when $x^2 = 400$ is easy?#

The ques­tion asks whether the design is pos­si­ble, and the dis­crim­i­nant answers that directly. Solv­ing $x^2 = 400$ then gives the dimen­sions.

What does $D \lt 0$ mean in a word prob­lem?#

It means no real value of the unknown sat­is­fies all the con­di­tions, so the sit­u­a­tion described can­not hap­pen, as in Ques­tion 4.

Why is the answer to Ques­tion 5 a square?#

With $D = 0$ there is only one pos­si­ble length, 20 m, and the breadth is then also 20 m. A square is a spe­cial rec­tan­gle, so it is a valid answer.

Ref­er­ences#

1. National Coun­cil of Edu­ca­tional Research and Train­ing. Math­e­mat­ics: Text­book for Class X. NCERT, New Delhi.
2. Sharma, R. D. Math­e­mat­ics for Class 10. Dhan­pat Rai Pub­li­ca­tions.
3. Aggar­wal, R. S. Sec­ondary School Math­e­mat­ics for Class 10. Bharati Bhawan.