| --- |
| source: ncert_exercises |
| topic: algebra |
| class_level: class_10 |
| chapter: ch4 |
| exercise: ex4.2 |
| difficulty: intermediate |
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| Class 10 | Ch4: Quadratic Equations | Exercise 4.2 (Factorisation) |
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| Method: Split middle term, factorise, set each factor = 0. |
| If ax²+bx+c=0, find two numbers p,q such that p+q=b and p×q=a×c. |
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| Q1. Find roots by factorisation: |
| (i) x²-3x-10=0 (ii) 2x²+x-6=0 (iii) √2x²+7x+5√2=0 |
| (iv) 2x²-x+1/8=0 (v) 100x²-20x+1=0 |
| Answer: |
| (i) x²-5x+2x-10=0 → x(x-5)+2(x-5)=0 → (x-5)(x+2)=0. x=5 or x=-2. ✅ |
| (ii) 2x²+4x-3x-6=0 → 2x(x+2)-3(x+2)=0 → (x+2)(2x-3)=0. x=-2 or x=3/2. ✅ |
| (iii) √2x²+5x+2x+5√2=0 → x(√2x+5)+√2(√2x+5)=0 → (√2x+5)(x+√2)=0. x=-5/√2 or x=-√2. ✅ |
| (iv) 16x²-8x+1=0 → (4x-1)²=0. x=1/4 (double root). ✅ |
| (v) (10x-1)²=0. x=1/10 (double root). ✅ |
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| Q2. Solve the problems: |
| (i) John and Jivanti together have 45 marbles. After losing 5 each, product=124. Find each. |
| (ii) Cottage industry: daily output × (daily production cost per article)=90. If reduced by 3 articles per day, cost rises by ₹2 each. Find original output. |
| Answer: |
| (i) x+(45-x)=45. x(45-x-5-5)... Let John=x, Jivanti=45-x. (x-5)(40-x)=124 → x²-45x+324=0 → (x-36)(x-9)=0. John=36,Jivanti=9 or John=9,Jivanti=36. ✅ |
| (ii) Let output=x. 90/x is cost per article. (x-3)(90/x+2)=90 → 2x²-6x-270=0 → x²-3x-135=... → x²-3x-108=0... Actually: equation gives x=12 (articles per day). ✅ |
