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Quadratic Equations

NCERT Class 10 Exercise 4.3 Solutions: Nature of Roots

Full NCERT Exercise 4.3 solutions using the discriminant: nature of roots of three equations, values of k for equal roots, and whether the mango grove, friends' ages and rectangular park situations are possible.

By 7 min read

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 ax2+bx+c=0ax^2 + bx + c = 0 with a≠0a \neq 0, the dis­crim­i­nant is D=b2−4acD = b^2 - 4ac, and

  • D>0D \gt 0: two dis­tinct real roots, x=−b±D2a\displaystyle x = \frac{-b \pm \sqrt{D}}{2a};
  • D=0D = 0: two equal real roots, each x=−b2a\displaystyle x = -\frac{b}{2a};
  • D<0D \lt 0: no real roots.
  1. Write the equa­tion in stan­dard form and read off aa, bb, cc with their signs.
  2. Com­pute DD and decide the nature of the roots.
  3. If the roots are real, find them.
  4. In a word prob­lem, a neg­a­tive DD 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)

2x2−3x+5=02x^2 - 3x + 5 = 0

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

D=b2−4ac=(−3)2−4(2)(5)=9−40=−31D = b^2 - 4ac = (-3)^2 - 4(2)(5) = 9 - 40 = -31

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

Why this works: com­plet­ing the square gives 2x2−3x+5=2(x−34)2+318\displaystyle 2x^2 - 3x + 5 = 2\left(x - \frac{3}{4}\right)^2 + \frac{31}{8}, which is at least 318\displaystyle \frac{31}{8} for every real xx, so it can never be 0.

Answer: no real roots.

Ques­tion 1 (ii)

3x2−43 x+4=03x^2 - 4\sqrt{3}\,x + 4 = 0

Solu­tion. Here a=3a = 3, b=−43b = -4\sqrt{3}, c=4c = 4.

D=(−43)2−4(3)(4)=16×3−48=48−48=0D = \left(-4\sqrt{3}\right)^2 - 4(3)(4) = 16 \times 3 - 48 = 48 - 48 = 0

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

x=−b2a=432×3=233=23\displaystyle x = -\frac{b}{2a} = \frac{4\sqrt{3}}{2 \times 3} = \frac{2\sqrt{3}}{3} = \frac{2}{\sqrt{3}}

Check: with x=233\displaystyle x = \frac{2\sqrt{3}}{3}, x2=129=43\displaystyle x^2 = \frac{12}{9} = \frac{4}{3}, so 3x2=43x^2 = 4; and 43 x=8×33=8\displaystyle 4\sqrt{3}\,x = \frac{8 \times 3}{3} = 8. Then 4−8+4=04 - 8 + 4 = 0.

Answer: two equal real roots, 23\displaystyle \frac{2}{\sqrt{3}} and 23\displaystyle \frac{2}{\sqrt{3}} (that is, 233\displaystyle \frac{2\sqrt{3}}{3} each).

Ques­tion 1 (iii)

2x2−6x+3=02x^2 - 6x + 3 = 0

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

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

Since D>0D \gt 0, there are two dis­tinct real roots. As 12 is not a per­fect square, they are irra­tional. Using 12=23\sqrt{12} = 2\sqrt{3}:

x=−b±D2a=6±234=3±32\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 3+32+3−32=3=−ba\displaystyle \frac{3 + \sqrt{3}}{2} + \frac{3 - \sqrt{3}}{2} = 3 = -\frac{b}{a}, and the prod­uct is (3+3)(3−3)4=9−34=32=ca\displaystyle \frac{(3 + \sqrt{3})(3 - \sqrt{3})}{4} = \frac{9 - 3}{4} = \frac{3}{2} = \frac{c}{a}.

Answer: two dis­tinct real roots, x=3+32\displaystyle x = \frac{3 + \sqrt{3}}{2} and x=3−32\displaystyle x = \frac{3 - \sqrt{3}}{2} (about 2.37 and 0.63, tak­ing 3≈1.732\sqrt{3} \approx 1.732).

Graphs of the three equations of Question 1: 2x squared - 3x + 5 stays above the x-axis; 3x squared - 4 root 3 x + 4 touches it at about 1.15; 2x squared - 6x + 3 crosses it at about 0.63 and 2.37.
Ques­tion 1: D=−31D = -31 (no cross­ing), D=0D = 0 (touch­ing at 233\displaystyle \frac{2\sqrt{3}}{3}) and D=12D = 12 (two cross­ings at 3±32\displaystyle \frac{3 \pm \sqrt{3}}{2}).

Ques­tion 2

Find the val­ues of kk for each of the fol­low­ing qua­dratic equa­tions, so that they have two equal roots.

Ques­tion 2 (i)

2x2+kx+3=02x^2 + kx + 3 = 0

Solu­tion. Here a=2a = 2, b=kb = k, c=3c = 3. For two equal roots, D=0D = 0:

k2−4(2)(3)=0k2=24k=±24=±26\begin{aligned} k^2 - 4(2)(3) &= 0 \\ k^2 &= 24 \\ k &= \pm\sqrt{24} = \pm 2\sqrt{6} \end{aligned}

Check: for both val­ues, k2=24k^2 = 24, so D=24−24=0D = 24 - 24 = 0.

Answer: k=26k = 2\sqrt{6} or k=−26k = -2\sqrt{6}.

Ques­tion 2 (ii)

kx(x−2)+6=0kx(x - 2) + 6 = 0

Solu­tion. First write it in stan­dard form:

kx2−2kx+6=0kx^2 - 2kx + 6 = 0

Here a=ka = k, b=−2kb = -2k, c=6c = 6. For the equa­tion to be qua­dratic we need k≠0k \neq 0. For two equal roots, D=0D = 0:

(−2k)2−4(k)(6)=04k2−24k=04k(k−6)=0\begin{aligned} (-2k)^2 - 4(k)(6) &= 0 \\ 4k^2 - 24k &= 0 \\ 4k(k - 6) &= 0 \end{aligned}

So k=0k = 0 or k=6k = 6. But k=0k = 0 makes the equa­tion 6=06 = 0, which is not a qua­dratic equa­tion (and is false), so we reject it.

Check: with k=6k = 6, the equa­tion is 6x2−12x+6=06x^2 - 12x + 6 = 0, that is, 6(x−1)2=06(x - 1)^2 = 0, with equal roots 1 and 1.

Answer: k=6k = 6.

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 xx m. Then the length is 2x2x m, and the area con­di­tion gives

2x×x=800  ⟹  2x2−800=02x \times x = 800 \implies 2x^2 - 800 = 0

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

D=02−4(2)(−800)=6400>0D = 0^2 - 4(2)(-800) = 6400 \gt 0

Real roots exist, so the design is pos­si­ble. To find them, 2x2=8002x^2 = 800 gives x2=400x^2 = 400, so x=20x = 20 or x=−20x = -20. A breadth can­not be neg­a­tive, so x=20x = 20.

Check: 40×20=80040 \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 xx years. Then the other is (20−x)(20 - x) years old. Four years ago their ages were (x−4)(x - 4) and (16−x)(16 - x) years, so

(x−4)(16−x)=4816x−x2−64+4x=48−x2+20x−112=0x2−20x+112=0\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=1a = 1, b=−20b = -20, c=112c = 112:

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

Since D<0D \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(x - 4) + (16 - x) = 12. Two num­bers with sum 12 have the largest prod­uct when both are 6, and 6×6=366 \times 6 = 36. Since 48 is more than 36, the prod­uct 48 can never be reached.

Answer: no, the sit­u­a­tion is not pos­si­ble.

Graph of the product (x - 4)(16 - x) for present ages x from 4 to 16, peaking at 36 when x = 10, below a dashed horizontal line at the required product 48.
Ques­tion 4: the prod­uct of the ages four years ago is at most 36, so it can never equal 48.

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 xx m. Since the perime­ter is 2(length+breadth)=802(\text{length} + \text{breadth}) = 80, length + breadth = 40, so the breadth is (40−x)(40 - x) m. The area con­di­tion gives

x(40−x)=40040x−x2=400x2−40x+400=0\begin{aligned} x(40 - x) &= 400 \\ 40x - x^2 &= 400 \\ x^2 - 40x + 400 &= 0 \end{aligned}

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

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

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

x=−b2a=402=20\displaystyle x = -\frac{b}{2a} = \frac{40}{2} = 20

So the length is 20 m and the breadth is 40−20=2040 - 20 = 20 m. The park is in fact a square.

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

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

Key terms

Dis­crim­i­nant
D=b2−4acD = b^2 - 4ac for the equa­tion ax2+bx+c=0ax^2 + bx + c = 0.
Nature of roots
Whether the roots are real and dis­tinct (D>0D \gt 0), real and equal (D=0D = 0), or not real (D<0D \lt 0).
Equal roots
Two iden­ti­cal real roots, each equal to −b2a\displaystyle -\frac{b}{2a}.
Irra­tional roots
Real roots that involve a surd, as when DD is pos­i­tive but not a per­fect square.
Qua­dratic for­mula
x=−b±b2−4ac2a\displaystyle x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, valid when b2−4ac≥0b^2 - 4ac \geq 0.
Sum and prod­uct of roots
For ax2+bx+c=0ax^2 + bx + c = 0, the roots add to −ba\displaystyle -\frac{b}{a} and mul­ti­ply to ca\displaystyle \frac{c}{a}; a quick check on answers.
Perime­ter
The total bound­ary length of a rec­tan­gle, 2(length+breadth)2(\text{length} + \text{breadth}).

Com­mon ques­tions

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

With k=0k = 0 the x2x^2 term van­ishes and the equa­tion becomes 6=06 = 0, which is not a qua­dratic equa­tion at all. The con­di­tion for equal roots applies only to qua­drat­ics.

Is 23\displaystyle \frac{2}{\sqrt{3}} the same as 233\displaystyle \frac{2\sqrt{3}}{3}?

Yes. Mul­ti­ply­ing the numer­a­tor and denom­i­na­tor of 23\displaystyle \frac{2}{\sqrt{3}} by 3\sqrt{3} gives 233\displaystyle \frac{2\sqrt{3}}{3}. Either form is accepted.

In Ques­tion 3, why use the dis­crim­i­nant when x2=400x^2 = 400 is easy?

The ques­tion asks whether the design is pos­si­ble, and the dis­crim­i­nant answers that directly. Solv­ing x2=400x^2 = 400 then gives the dimen­sions.

What does D<0D \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=0D = 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.
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