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NCERT Solutions · Class 10 · Chapter 4: Quadratic Equations

NCERT Solutions for Class 10 Maths Chapter 4 Exercise 4.3

Exercise 4.3: Nature of roots (the discriminant). For \(ax^2 + bx + c = 0\), the discriminant is \(D = b^2 - 4ac\). \(D > 0\): two distinct real roots; \(D = 0\): two equal real roots \(x = -\dfrac{b}{2a}\); \(D < 0\): no real roots. The roots are \(x = \dfrac{-b \pm \sqrt{D}}{2a}\).

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Exercise 4.3 questions and solutions

Exercise 4.3, Question 1

Find the nature of the roots; if real roots exist, find them.
(i) \(2x^2 - 3x + 5 = 0\)
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  1. \(a = 2, b = -3, c = 5\): \(D = (-3)^2 - 4(2)(5) = 9 - 40 = -31\).
  2. \(D < 0\).
Answer: No real roots
(ii) \(3x^2 - 4\sqrt{3}\,x + 4 = 0\)
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  1. \[\begin{aligned}D &= (-4\sqrt{3})^2 - 4(3)(4) \\ &= 48 - 48 \\ &= 0\end{aligned}\]: two equal real roots.
  2. \[\begin{aligned}x &= -\dfrac{b}{2a} \\ &= \dfrac{4\sqrt{3}}{6} \\ &= \dfrac{2\sqrt{3}}{3} \\ &= \dfrac{2}{\sqrt{3}}\end{aligned}\]
Answer: Equal roots: \[x = \dfrac{2}{\sqrt{3}}, \dfrac{2}{\sqrt{3}}\]
(iii) \(2x^2 - 6x + 3 = 0\)
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  1. \(D = (-6)^2 - 4(2)(3) = 36 - 24 = 12 > 0\): two distinct real roots.
  2. \[\begin{aligned}x &= \dfrac{6 \pm \sqrt{12}}{4} \\ &= \dfrac{6 \pm 2\sqrt{3}}{4} \\ &= \dfrac{3 \pm \sqrt{3}}{2}\end{aligned}\]
Answer: Distinct real roots: \[x = \dfrac{3 + \sqrt{3}}{2}, \dfrac{3 - \sqrt{3}}{2}\]

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Exercise 4.3, Question 2

Find the values of \(k\) for which each equation has two equal roots.
(i) \(2x^2 + kx + 3 = 0\)
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  1. Equal roots when \(D = 0\): \(k^2 - 4(2)(3) = 0 \Rightarrow k^2 = 24\).
  2. \(k = \pm\sqrt{24} = \pm 2\sqrt{6}\).
Answer: \(k = 2\sqrt{6}\) or \(k = -2\sqrt{6}\)
(ii) \(kx(x - 2) + 6 = 0\)
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  1. Standard form: \(kx^2 - 2kx + 6 = 0\), so \(a = k, b = -2k, c = 6\).
  2. \(D = 4k^2 - 24k = 4k(k - 6)\). Setting \(D = 0\) gives \(k = 0\) or \(k = 6\).
  3. \(k = 0\) gives \(6 = 0\), which is not a quadratic at all, so reject it.
Answer: \(k = 6\)

Where marks slip: Always check that the value of \(k\) keeps the \(x^2\) coefficient non-zero.

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Exercise 4.3, Question 3

Is it possible to lay out a rectangular grove whose length is twice its breadth and whose area is \(800\ \text{m}^2\)? If so, find its length and breadth.
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  1. Let the breadth be \(x\) m, length \(2x\) m: \(2x^2 = 800 \Rightarrow x^2 - 400 = 0\).
  2. \(D = 0 - 4(1)(-400) = 1600 > 0\), so real roots exist: \(x = \pm 20\). Take \(x = 20\) (a length is positive).
Answer: Yes: breadth \(20\) m, length \(40\) m

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Exercise 4.3, Question 4

Is this possible? The ages of two friends add up to \(20\) years, and four years ago the product of their ages was \(48\).
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  1. Let one age be \(x\); the other is \(20 - x\). Four years ago: \((x - 4)(16 - x) = 48\).
  2. \[\begin{aligned}&-x^2 + 20x - 64 = 48 \\ \Rightarrow\ &x^2 - 20x + 112 = 0\end{aligned}\]
  3. \(D = 400 - 448 = -48 < 0\): no real roots.
Answer: Not possible: the equation has no real solution.

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Exercise 4.3, Question 5

Is it possible to design a rectangular park with perimeter \(80\) m and area \(400\ \text{m}^2\)? If so, find its length and breadth.
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  1. Perimeter \(80\) means length + breadth \(= 40\). Let the length be \(x\); breadth \(40 - x\).
  2. \[\begin{aligned}&x(40 - x) = 400 \\ \Rightarrow\ &x^2 - 40x + 400 = 0\end{aligned}\]
  3. \(D = 1600 - 1600 = 0\): equal roots, \(x = 20\). So length \(= 20\) and breadth \(= 20\).
Answer: Yes: \(20\) m by \(20\) m (a square)

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Textbook: NCERT Mathematics Class 10 (rationalised edition, 2023-24 reprint onward), free from ncert.nic.in. Question statements are shortened to the minimum needed; the solutions and tips are our own. CBSE Math Revision is independent and not affiliated with NCERT or CBSE. Spotted a slip? Tell us and it goes in the corrections log.