NCERT Mathematics · Class 10 · English · 2026-27
Quadratic Equations — Exercise 4.2
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
Find the roots of the five quadratics in Question 1 by factorisation.
Step-by-step solution
Rewrite each equation as a product equal to zero, then set each factor equal to zero.
(i) x²−3x−10=0
\[x^2-5x+2x-10=0\]
\[(x-5)(x+2)=0\]
(ii) 2x²+x−6=0
\[2x^2+4x-3x-6=0\]
\[(2x-3)(x+2)=0\]
(iii) √2 x²+7x+5√2=0
\[\sqrt2x^2+2x+5x+5\sqrt2=0\]
\[(\sqrt2x+5)(x+\sqrt2)=0\]
(iv) 2x²−x+1/8=0
Multiply by 8 to remove the fraction:
\[16x^2-8x+1=(4x-1)^2=0\]
(v) 100x²−20x+1=0
\[(10x-1)^2=0\]
Resource ID: CB-10-M-CH4-EX4.2-Q1
Question 2
Solve the two situations from Example 1: the original marble counts and the possible daily toy production.
Step-by-step solution
(i) John and Jivanti’s marbles
Let John start with x marbles; Jivanti starts with 45−x. After each loses five, the counts are x−5 and 40−x.
\[(x-5)(40-x)=124\]
\[-x^2+45x-200=124\;\Rightarrow\;x^2-45x+324=0\]
\[(x-36)(x-9)=0\]
(ii) Toys made in one day
Let n be the number of toys. Cost per toy is ₹(55−n), and total cost is ₹750.
\[n(55-n)=750\;\Rightarrow\;n^2-55n+750=0\]
\[(n-25)(n-30)=0\]
Resource ID: CB-10-M-CH4-EX4.2-Q2
Question 3
Find two numbers with sum 27 and product 182.
Step-by-step solution
If one number is x, the other is 27−x. Their product gives:
\[x(27-x)=182\;\Rightarrow\;x^2-27x+182=0\]
\[(x-13)(x-14)=0\]
Resource ID: CB-10-M-CH4-EX4.2-Q3
Question 4
Find consecutive positive integers whose squares add to 365.
Step-by-step solution
Let the smaller integer be n. The next is n+1.
\[n^2+(n+1)^2=365\]
\[2n^2+2n-364=0\;\Rightarrow\;n^2+n-182=0\]
\[(n+14)(n-13)=0\]
The root n=−14 is excluded because the integers must be positive.
Resource ID: CB-10-M-CH4-EX4.2-Q4
Question 5
A right triangle has hypotenuse 13 cm and altitude 7 cm shorter than its base. Find the other two sides.
Step-by-step solution
Let the base be b cm, so the altitude is b−7 cm and b>7. Apply Pythagoras:
\[b^2+(b-7)^2=13^2\]
\[2b^2-14b-120=0\;\Rightarrow\;b^2-7b-60=0\]
\[(b-12)(b+5)=0\]
Reject b=−5 because a length cannot be negative. Thus b=12 and altitude=12−7=5.
Resource ID: CB-10-M-CH4-EX4.2-Q5
Question 6
A pottery batch costs ₹90 in total; the cost per article is ₹(2n+3) when n articles are made. Find n and the unit cost.
Step-by-step solution
Count×unit cost equals the total cost:
\[n(2n+3)=90\;\Rightarrow\;2n^2+3n-90=0\]
\[2n^2+15n-12n-90=0\]
\[(2n+15)(n-6)=0\]
The roots are n=6 and n=−15/2. A count must be a positive integer, so only n=6 is valid.
Resource ID: CB-10-M-CH4-EX4.2-Q6