Examples Quadratic equations class 10 | Quadratic equations examples solution class 10

Examples Quadratic equations class 10 | Quadratic equations examples solution class 10 by @rajansir07 #QuadraticEquations #Class10Maths #MathsWithRajanSir #CBSEMaths #BoardExamPreparation #QuadraticFormula #Class10CBSE #MathsExamples #Class10Board #MathsSolutions #CBSE2025 #QuadraticEquationsExplained #EasyMathsTricks #MathsMadeEasy #NCERTMaths Example 1 : Represent the following situations mathematically: (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. We would like to 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 rupees) was found to be 55 minus the number of toys produced in a day. On a particular day, the total cost of production was ` 750. We would like to find out the number of toys produced on that day. Example 2 : Check whether the following are quadratic equations: (i) (x – 2)2 + 1 = 2x – 3 (ii) x(x + 1) + 8 = (x + 2) (x – 2) (iii) x (2x + 3) = x2 + 1 (iv) (x + 2)3 = x3 – 4 Example 3 : Find the roots of the equation 2x2 – 5x + 3 = 0, by factorisation. Example 4 : Find the roots of the quadratic equation 6x2 – x – 2 = 0 Example 5 : Find the roots of the quadratic equation 2 x x 3 2 6 2 0 Example 6 : Find the dimensions of the prayer hall discussed in Section 4.1. Solution : In Section 4.1, we found that if the breadth of the hall is x m, then x satisfies the equation 2x2 + x – 300 = 0. Applying the factorisation method, we write this equation as 2x2 – 24x + 25x – 300 = 0 2x (x – 12) + 25 (x – 12) = 0 i.e., (x – 12)(2x + 25) = 0 So, the roots of the given equation are x = 12 or x = – 12.5. Since x is the breadth of the hall, it cannot be negative. Thus, the breadth of the hall is 12 m. Its length = 2x + 1 = 25 m. Example 7: Find the discriminant of the quadratic equation 2x2 – 4x + 3 = 0, and hence find the nature of its roots. Example 8 : A pole has to be erected at a point on the boundary of a circular park of diameter 13 metres in such a way that the differences of its distances from two diametrically opposite fixed gates A and B on the boundary is 7 metres. Is it possible to do so? If yes, at what distances from the two gates should the pole be erected? Solution : Let us first draw the diagram (see Fig. 4.2). Let P be the required location of the pole. Let the distance of the pole from the gate B be x m, i.e., BP = x m. Now the difference of the distances of the pole from the two gates = AP – BP (or, BP – AP) = 7 m. Therefore, AP = (x + 7) m. Now, AB = 13m, and since AB is a diameter, APB = 90° Therefore, i.e., i.e., Example 9 : Find the discriminant of the equation 3x2 – 2x + 1 3 nature of its roots. Find them, if they are real. 1 3 Solution : Here a = 3, b = – 2 and c  . Therefore, discriminant b2 – 4ac = (– 2)2 – 4 × 3 × 1 3 = 0 and hence find the = 4 – 4 = 0. Hence, the given quadratic equation has two equal real roots. The roots are   b b a a i.e., 2 2 1 1 , , , i.e., , , . 2 2 6 6 3 3