GSEB Class 10 Maths Solutions Chapter 4 દ્વિઘાત સમીકરણ Exercise 4.2

Question 1. Find the roots of the following quadratic equations by factorization method:
(i) \( x^2 - 3x - 10 = 0 \)
(ii) \( 2x^2 + x - 6 = 0 \)
(iii) \( \sqrt{2}x^2 + 7x + 5\sqrt{2} = 0 \)
(iv) \( \frac{16}{x} - 1 = \frac{15}{x+1}; x \neq 0, -1 \)
(v) \( 100x^2 - 20x + 1 = 0 \)
Answer:
(i) \( x^2 - 5x + 2x - 10 = 0 \)
\( x(x - 5) + 2(x - 5) = 0 \)
\( (x - 5)(x + 2) = 0 \)
\( x = 5 \) or \( x = -2 \)
(ii) \( 2x^2 + 4x - 3x - 6 = 0 \)
\( 2x(x + 2) - 3(x + 2) = 0 \)
\( (x + 2)(2x - 3) = 0 \)
\( x = -2 \) or \( x = \frac{3}{2} \)
(iii) \( \sqrt{2}x^2 + 5x + 2x + 5\sqrt{2} = 0 \)
\( x(\sqrt{2}x + 5) + \sqrt{2}(\sqrt{2}x + 5) = 0 \)
\( (\sqrt{2}x + 5)(x + \sqrt{2}) = 0 \)
\( x = -\frac{5}{\sqrt{2}} \) or \( x = -\sqrt{2} \)
(iv) \( \frac{16(x+1) - 15x}{x(x+1)} = 1 \)
\( 16x + 16 - 15x = x^2 + x \)
\( x^2 = 16 \)
\( x = \pm 4 \)
(v) \( 100x^2 - 10x - 10x + 1 = 0 \)
\( 10x(10x - 1) - 1(10x - 1) = 0 \)
\( (10x - 1)^2 = 0 \)
\( x = \frac{1}{10} \)
In simple words: To find the roots, split the middle term to factor the equation into two parts. Set each part to zero to solve for x.

Exam Tip: Always check your factors by expanding them back to ensure they match the original equation before solving for x.

Question 2. (i) John and Jivanti together have 45 marbles. Both of them lost 5 marbles each, and the product of the number of marbles they now have is 124. Find out how many marbles they had to start with. (ii) A cottage industry produces a certain number of toys in a day. The cost of production of each toy (in Rs.) was found to be 55 minus the number of articles produced in a day. On a particular day, the total cost of production was 750. Find out the number of toys produced on that day.
Answer:
(i) Let John have \( x \) marbles. Jivanti has \( 45 - x \). After losing 5 each: \( (x-5)(45-x-5) = 124 \).
\( (x-5)(40-x) = 124 \)
\( -x^2 + 45x - 200 = 124 \)
\( x^2 - 45x + 324 = 0 \)
\( (x-36)(x-9) = 0 \). So, John had 36 or 9 marbles, and Jivanti had 9 or 36 marbles.
(ii) Let toys produced be \( x \). Cost per toy is \( 55 - x \). Total cost \( x(55-x) = 750 \).
\( 55x - x^2 = 750 \)
\( x^2 - 55x + 750 = 0 \)
\( (x-30)(x-25) = 0 \). So, 30 or 25 toys were produced.
In simple words: Turn the word problem into an equation by assigning 'x' to the unknown value. Solve the resulting quadratic equation to find the answer.

Exam Tip: Read the problem carefully to identify the relationship between the variables, as this is key to forming the correct quadratic equation.

Question 3. Find two numbers whose sum is 27, and product is 182.
Answer: Let the first number be \( x \). The second number is \( 27 - x \).
\( x(27 - x) = 182 \)
\( 27x - x^2 = 182 \)
\( x^2 - 27x + 182 = 0 \)
\( x^2 - 14x - 13x + 182 = 0 \)
\( x(x - 14) - 13(x - 14) = 0 \)
\( (x - 13)(x - 14) = 0 \)
\( x = 13 \) or \( x = 14 \). The numbers are 13 and 14.
In simple words: If two numbers add up to 27, call them x and 27-x. Multiply them to get 182 and solve the equation.

Exam Tip: When you have the sum and product, the numbers are roots of the quadratic equation \( t^2 - (\text{sum})t + (\text{product}) = 0 \).

Question 4. Find two consecutive positive integers, sum of whose square is 365.
Answer: Let the integers be \( x \) and \( x + 1 \).
\( x^2 + (x + 1)^2 = 365 \)
\( x^2 + x^2 + 2x + 1 = 365 \)
\( 2x^2 + 2x - 364 = 0 \)
\( x^2 + x - 182 = 0 \)
\( x^2 + 14x - 13x - 182 = 0 \)
\( x(x + 14) - 13(x + 14) = 0 \)
\( (x - 13)(x + 14) = 0 \). Since the integer is positive, \( x = 13 \). The integers are 13 and 14.
In simple words: Consecutive integers are x and x+1. Square them, add them, and solve the equation to find the numbers.

Exam Tip: Always discard negative solutions if the question specifically asks for positive integers.

Question 5. The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm. Find the other two sides.
Answer: Let base \( = x \) cm, altitude \( = x - 7 \) cm, hypotenuse \( = 13 \) cm.
By Pythagoras theorem: \( x^2 + (x - 7)^2 = 13^2 \)
\( x^2 + x^2 - 14x + 49 = 169 \)
\( 2x^2 - 14x - 120 = 0 \)
\( x^2 - 7x - 60 = 0 \)
\( x^2 - 12x + 5x - 60 = 0 \)
\( x(x - 12) + 5(x - 12) = 0 \)
\( (x + 5)(x - 12) = 0 \). Since side cannot be negative, \( x = 12 \). Base is 12 cm, altitude is 5 cm.
In simple words: Use the Pythagorean theorem (base squared plus height squared equals hypotenuse squared) to create an equation and solve for the base.

Exam Tip: Remember that in geometry problems, lengths must always be positive, so reject any negative values for x.

Question 6. A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that the cost of production of each article (in rupees) was 3 more than twice the number of articles produces on that day. If the total cost of production on that day was 90. Find the number of articles produced and cost of each articles.
Answer: Let articles produced be \( x \). Cost per article is \( 2x + 3 \).
Total cost \( = x(2x + 3) = 90 \)
\( 2x^2 + 3x - 90 = 0 \)
\( 2x^2 + 15x - 12x - 90 = 0 \)
\( x(2x + 15) - 6(2x + 15) = 0 \)
\( (x - 6)(2x + 15) = 0 \). Since \( x \) cannot be negative, \( x = 6 \).
Number of articles is 6, cost per article is \( 2(6) + 3 = 15 \) Rs.
In simple words: Multiply the number of items by the cost per item to get the total cost, then solve the quadratic equation.

Exam Tip: Clearly define your variables at the start to avoid confusion when setting up the total cost equation.