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NCERT Solutions · Class 10 · Chapter 4: Quadratic Equations

NCERT Solutions for Class 10 Maths Chapter 4 Exercise 4.1

Exercise 4.1: Is it a quadratic equation? Forming quadratics. An equation is quadratic when, after expanding and simplifying, it can be written as \(ax^2 + bx + c = 0\) with \(a \ne 0\). Always simplify first: an \(x^2\) term on both sides can cancel, and a cubic can reduce to a quadratic.

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Exercise 4.1 questions and solutions

Exercise 4.1, Question 1

Decide whether each equation is quadratic.
(i) \((x + 1)^2 = 2(x - 3)\)
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  1. Expand and bring every term to one side: \[\begin{aligned}&x^2 + 2x + 1 - 2x + 6 = 0 \\ \Rightarrow\ &x^2 + 7 = 0\end{aligned}\]
Answer: Yes: \(x^2 + 7 = 0\)
(ii) \(x^2 - 2x = (-2)(3 - x)\)
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  1. Expand and bring every term to one side: \[\begin{aligned}&x^2 - 2x + 6 - 2x = 0 \\ \Rightarrow\ &x^2 - 4x + 6 = 0\end{aligned}\]
Answer: Yes: \(x^2 - 4x + 6 = 0\)
(iii) \((x - 2)(x + 1) = (x - 1)(x + 3)\)
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  1. Expand and bring every term to one side: \[\begin{aligned}&x^2 - x - 2 = x^2 + 2x - 3 \\ \Rightarrow\ &-3x + 1 = 0\end{aligned}\]
Answer: No: the \(x^2\) terms cancel, leaving \(-3x + 1 = 0\) (linear)
(iv) \((x - 3)(2x + 1) = x(x + 5)\)
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  1. Expand and bring every term to one side: \[\begin{aligned}&2x^2 - 5x - 3 = x^2 + 5x \\ \Rightarrow\ &x^2 - 10x - 3 = 0\end{aligned}\]
Answer: Yes: \(x^2 - 10x - 3 = 0\)
(v) \((2x - 1)(x - 3) = (x + 5)(x - 1)\)
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  1. Expand and bring every term to one side: \[\begin{aligned}&2x^2 - 7x + 3 = x^2 + 4x - 5 \\ \Rightarrow\ &x^2 - 11x + 8 = 0\end{aligned}\]
Answer: Yes: \(x^2 - 11x + 8 = 0\)
(vi) \(x^2 + 3x + 1 = (x - 2)^2\)
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  1. Expand and bring every term to one side: \[\begin{aligned}&x^2 + 3x + 1 = x^2 - 4x + 4 \\ \Rightarrow\ &7x - 3 = 0\end{aligned}\]
Answer: No: \(7x - 3 = 0\) (linear)
(vii) \((x + 2)^3 = 2x(x^2 - 1)\)
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  1. Expand and bring every term to one side: \[\begin{aligned}&x^3 + 6x^2 + 12x + 8 = 2x^3 - 2x \\ \Rightarrow\ &x^3 - 6x^2 - 14x - 8 = 0\end{aligned}\]
Answer: No: it is a cubic, \(x^3 - 6x^2 - 14x - 8 = 0\)
(viii) \(x^3 - 4x^2 - x + 1 = (x - 2)^3\)
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  1. Expand and bring every term to one side: \[\begin{aligned}&x^3 - 4x^2 - x + 1 = x^3 - 6x^2 + 12x - 8 \\ \Rightarrow\ &2x^2 - 13x + 9 = 0\end{aligned}\]
Answer: Yes: the \(x^3\) terms cancel, leaving \(2x^2 - 13x + 9 = 0\)

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Exercise 4.1, Question 2

Write each situation as a quadratic equation. (Only the equation is asked for; the root is shown as a check.)
(i) A rectangular plot has area \(528\ \text{m}^2\); its length is one more than twice its breadth.
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  1. Let the breadth be \(x\) m. Then the length is \((2x + 1)\) m.
  2. Area: \[\begin{aligned}&x(2x + 1) = 528 \\ \Rightarrow\ &2x^2 + x - 528 = 0\end{aligned}\]
  3. Check: \(2x^2 + x - 528 = (x - 16)(2x + 33)\), so \(x = 16\): breadth \(16\) m, length \(33\) m.
Answer: \(2x^2 + x - 528 = 0\) (breadth \(x\) m)
(ii) The product of two consecutive positive integers is \(306\).
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  1. Let the integers be \(x\) and \(x + 1\).
  2. \[\begin{aligned}&x(x + 1) = 306 \\ \Rightarrow\ &x^2 + x - 306 = 0\end{aligned}\]
  3. Check: \((x - 17)(x + 18) = 0\), so the integers are \(17\) and \(18\).
Answer: \(x^2 + x - 306 = 0\)
(iii) A mother is 26 years older than her son; in 3 years the product of their ages will be \(360\).
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  1. Let the son's age now be \(x\) years; the mother is \(x + 26\).
  2. In 3 years: \[\begin{aligned}&(x + 3)(x + 29) = 360 \\ \Rightarrow\ &x^2 + 32x + 87 - 360 = 0 \\ \Rightarrow\ &x^2 + 32x - 273 = 0\end{aligned}\]
  3. Check: \((x - 7)(x + 39) = 0\), so the son is \(7\) and the mother \(33\).
Answer: \(x^2 + 32x - 273 = 0\) (son's age \(x\))
(iv) A train covers \(480\) km at a uniform speed; 8 km/h slower, it would take 3 hours longer.
Show solution
  1. Let the speed be \(x\) km/h. Time taken \(= \dfrac{480}{x}\) hours.
  2. \(\dfrac{480}{x - 8} - \dfrac{480}{x} = 3\).
  3. Multiply by \(x(x - 8)\): \[\begin{aligned}&480x - 480(x - 8) = 3x(x - 8) \\ \Rightarrow\ &3840 = 3x^2 - 24x\end{aligned}\]
  4. Divide by 3: \(x^2 - 8x - 1280 = 0\). Check: \((x - 40)(x + 32) = 0\), so the speed is \(40\) km/h.
Answer: \(x^2 - 8x - 1280 = 0\) (speed \(x\) km/h)

Where marks slip: Time = distance ÷ speed. Write the slower journey's time minus the faster one's time, not the other way round.

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Quadratic Equations in our sample papers: Sample paper 1 (questions 4, 32) · Sample paper 2 (questions 4, 27, 37) · Sample paper 3 (questions 5, 22) · Sample paper 4 (questions 5, 22, 32, 37) · Sample paper 5 (questions 4, 5, 19, 27).

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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.