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Mathematics · Quadratic Equations · NCERT Exercises

Exercise 4.2

Complete, independently verified solutions for NCERT Exercise 4.2.

Mathematics · Chapter 4 · NCERT Exercises

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Question

Question 1

factorisation · roots

Find the roots by factorisation.

  1. x2−3x−10=0x^2-3x-10=0
  2. 2x2+x−6=02x^2+x-6=0
  3. 2x2+7x+52=0\sqrt2x^2+7x+5\sqrt2=0
  4. 2x2−x+18=02x^2-x+\frac18=0
  5. 100x2−20x+1=0100x^2-20x+1=0

Solution

  1. (x−5)(x+2)=0⇒x=5,−2(x-5)(x+2)=0\Rightarrow x=5,-2
  2. (2x−3)(x+2)=0⇒x=32,−2(2x-3)(x+2)=0\Rightarrow x=\frac32,-2
  3. (2x+5)(x+2)=0⇒x=−52,−2(\sqrt2x+5)(x+\sqrt2)=0\Rightarrow x=-\frac5{\sqrt2},-\sqrt2
  4. 16x2−8x+1=(4x−1)2=0⇒x=14,1416x^2-8x+1=(4x-1)^2=0\Rightarrow x=\frac14,\frac14
  5. (10x−1)2=0⇒x=110,110(10x-1)^2=0\Rightarrow x=\frac1{10},\frac1{10}

Substitution of every listed root in its original equation gives zero.

Question

Question 2

factorisation · word problems

Solve the two problems given in Example 1.

Solution

  1. x2−45x+324=(x−9)(x−36)=0x^2-45x+324=(x-9)(x-36)=0

    The initial marble counts are 9 and 36, in either order; after losing five each, 4 × 31 = 124.

  2. x2−55x+750=(x−25)(x−30)=0x^2-55x+750=(x-25)(x-30)=0

    The industry made either 25 toys at ₹30 each or 30 toys at ₹25 each; either total is ₹750.

Question

Question 3

factorisation · word problem

Find two numbers whose sum is 27 and product is 182.

Solution

x(27−x)=182⇒x2−27x+182=0x(27-x)=182\Rightarrow x^2-27x+182=0
(x−13)(x−14)=0(x-13)(x-14)=0

Both roots give the pair 13 and 14, whose sum and product verify the conditions.

Final answer

13 and 14.

Question

Question 4

factorisation · word problem

Find two consecutive positive integers whose squares sum to 365.

Solution

x2+(x+1)2=365⇒x2+x−182=0x^2+(x+1)^2=365\Rightarrow x^2+x-182=0
(x−13)(x+14)=0(x-13)(x+14)=0

x = −14 is rejected because the integers must be positive. For x = 13, 13² + 14² = 365.

Final answer

13 and 14.

Question

Question 5

factorisation · word problem · right triangle

The altitude of a right triangle is 7 cm less than its base and the hypotenuse is 13 cm. Find the other sides.

Solution

Let the base be x cm; altitude = x − 7 cm.

x2+(x−7)2=132⇒x2−7x−60=0x^2+(x-7)^2=13^2\Rightarrow x^2-7x-60=0
(x−12)(x+5)=0(x-12)(x+5)=0

x = −5 is impossible for a length. Hence the base is 12 cm and altitude is 5 cm; 5² + 12² = 13².

Final answer

Base = 12 cm; altitude = 5 cm.

Question

Question 6

factorisation · word problem

A cottage industry makes pottery articles. The cost per article is ₹3 more than twice the number made; total cost is ₹90. Find the number and cost per article.

Solution

Let x be the number of articles; cost per article = ₹(2x + 3).

x(2x+3)=90⇒2x2+3x−90=0x(2x+3)=90\Rightarrow2x^2+3x-90=0
(2x+15)(x−6)=0(2x+15)(x-6)=0

x = −7.5 is rejected because a production count cannot be negative. For x = 6, the cost is ₹15 and 6 × 15 = 90.

Final answer

6 articles; ₹15 per article.