NCERT Mathematics · Class 10 · English · 2026-27

Quadratic Equations — Exercise 4.1

Independently written, step-by-step solutions. Prompts below are concise paraphrases; use the official textbook for the complete question wording.

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Question 1

Expand and simplify the eight equations in Question 1 to decide which are quadratic.

Step-by-step solution

A quadratic equation reduces to ax²+bx+c=0 with a≠0. Expand both sides and collect like terms; the highest surviving power decides the degree.

(i) (x+1)²=2(x−3)

\[x^2+2x+1=2x-6\;\Rightarrow\;x^2+7=0\]

Answer: Quadratic. It need not have real roots to be a quadratic equation.

(ii) x²−2x=(−2)(3−x)

\[x^2-2x=-6+2x\;\Rightarrow\;x^2-4x+6=0\]

Answer: Quadratic.

(iii) (x−2)(x+1)=(x−1)(x+3)

\[x^2-x-2=x^2+2x-3\;\Rightarrow\;-3x+1=0\]

Answer: Not quadratic; the x² terms cancel and a linear equation remains.

(iv) (x−3)(2x+1)=x(x+5)

\[2x^2-5x-3=x^2+5x\;\Rightarrow\;x^2-10x-3=0\]

Answer: Quadratic.

(v) (2x−1)(x−3)=(x+5)(x−1)

\[2x^2-7x+3=x^2+4x-5\;\Rightarrow\;x^2-11x+8=0\]

Answer: Quadratic.

(vi) x²+3x+1=(x−2)²

\[x^2+3x+1=x^2-4x+4\;\Rightarrow\;7x-3=0\]

Answer: Not quadratic; it is linear.

(vii) (x+2)³=2x(x²−1)

\[x^3+6x^2+12x+8=2x^3-2x\]

\[-x^3+6x^2+14x+8=0\]

Answer: Not quadratic; it is cubic.

(viii) x³−4x²−x+1=(x−2)³

\[x^3-4x^2-x+1=x^3-6x^2+12x-8\]

\[2x^2-13x+9=0\]

Answer: Quadratic; the cubic terms cancel.

Resource ID: CB-10-M-CH4-EX4.1-Q1

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Question 2

Form quadratic equations for the rectangle, consecutive integers, ages and train-speed situations in Question 2.

Step-by-step solution

(i) Rectangle of area 528 m²

Let breadth be b metres. Length is 2b+1 metres. Since area is length×breadth:

\[b(2b+1)=528\;\Rightarrow\;2b^2+b-528=0,\quad b>0\]

Answer: 2b²+b−528=0.

(ii) Consecutive positive integers with product 306

Let the smaller integer be n; the next is n+1.

\[n(n+1)=306\;\Rightarrow\;n^2+n-306=0\]

Answer: n²+n−306=0, with n a positive integer.

(iii) Rohan and his mother

Let Rohan’s present age be r years. His mother is r+26. After three years their ages are r+3 and r+29.

\[(r+3)(r+29)=360\]

\[r^2+32r+87-360=0\;\Rightarrow\;r^2+32r-273=0\]

Answer: r²+32r−273=0, with r≥0.

(iv) Train travelling 480 km

Let normal speed be v km/h, where v>8. Time at normal speed is 480/v hours; at the lower speed it is 480/(v−8).

\[\frac{480}{v-8}-\frac{480}{v}=3\]

\[3840=3v(v-8)\;\Rightarrow\;v^2-8v-1280=0\]

Answer: v²−8v−1280=0, with v>8.

Resource ID: CB-10-M-CH4-EX4.1-Q2

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