There, we explained solving of the Quadratic Equations by Factoring and by Quadratic formula respectively with a number of solved Examples.

Here, a number of math word problems are worked out.

Several math word problems are given for practice in exercise with answers.

Practice makes one perfect.

This is especially true for learning of solving Math Word Problems.

So, take the exercises seriously and practice solving the Math Word Problems.

Example 1:

Solve the Following Math Word Problem.

The product of two consecutive odd integers is 899. Find them.

Solution to Example 1 of Math Word Problems :

We know, two consecutive odd integers differ by 2. Let the two consecutive odd integers be (x - 2) and x. By data, their product = 899 ⇒ (x - 2)x = 899 ⇒ x^{2} - 2x - 899 = 0 Comparing this equation with ax^{2} + bx + c = 0, we get a = 1, b = -2 and c = -899 We know by Quadratic Formula, x = {-b ± √(b^{2} - 4ac)}⁄2a Applying this Quadratic Formula here, we get x = {-b ± √(b^{2} - 4ac)}⁄2a = [-(-2) ± √{ (-2)^{2} - 4(1)( -899)}]⁄2(1)= [2 ± √{ 4 + 4(1)(899)}]⁄2 = [2 ± √{4(1 + 899)}]⁄2 = [2 ± √{4(900)}]⁄2 = [2 ± 2(30)}]⁄2 = 1±30 = 1+30 or 1-30 = 31 or -29 The two consecutive odd integers are (x - 2) and x. When x. = 31, the required integers are 31 - 2 and 31 i.e. 29 and 31. Ans. When x. = -29, the required integers are -29 - 2 and -29 i.e. -31 and -29. Ans.

Example 2 of Math Word Problems

Solve the Following Math Word Problem.

Find the number which is less than its square by 182.

Solution to Example 2 of Math Word Problems :

Let the required number be x. Then, by data, it is less than its square by 182. ⇒ x = x^{2} - 182 ⇒ x^{2} - x - 182 = 0 Comparing this equation with ax^{2} + bx + c = 0, we get a = 1, b = -1 and c = -182 We know by Quadratic Formula, x = {-b ± √(b^{2} - 4ac)}⁄2a Applying this Quadratic Formula here, we get x = {-b ± √(b^{2} - 4ac)}⁄2a = [-(-1) ± √{ (-1)^{2} - 4(1)( -182)}]⁄2(1)= [1 ± √{ 1 + 4(1)(182)}]⁄2 = [1 ± √{(1 + 728)}]⁄2 = [1 ± √{729}]⁄2 = [1 ± 27]⁄2 = (1 + 27)⁄2, (1 - 27)⁄2 = (28)⁄2, (- 26)⁄2 = 14, -13 Thus, the number which is less than its square by 182 is 14 or -13. Ans.

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The sum of the squares of two odd integers is 202. Find them.

Solution to Example 3 :

We know, two consecutive odd integers differ by 2. Let the two consecutive odd integers be (x - 2) and x. By data, the sum of their squares = 202 ⇒ (x - 2)^{2} + x^{2} = 202 ⇒ x^{2} - 4x + 4 + x^{2} = 202 ⇒ 2x^{2} - 4x + 4 = 202 ⇒ x^{2} - 2x + 2 = 101 ⇒ x^{2} - 2x - 99 = 0Comparing this equation with ax^{2} + bx + c = 0, we get a = 1, b = -2 and c = -99 We know by Quadratic Formula, x = {-b ± √(b^{2} - 4ac)}⁄2a Applying this Quadratic Formula here, we get x = {-b ± √(b^{2} - 4ac)}⁄2a = [-(-2) ± √{ (-2)^{2} - 4(1)( -99)}]⁄2(1)= [2 ± √{ 4 + 4(1)(99)}]⁄2 = [2 ± √{4(1 + 99)}]⁄2 = [2 ± √{4(100)}]⁄2 = [2 ± 2(10)}]⁄2 = 1±10 = 1+10 or 1-10 = 11 or -9 The two consecutive odd integers are (x - 2) and x. When x. = 11, the required integers are 11 - 2 and 11 i.e. 9 and 11. Ans. When x. = -9, the required integers are -9 - 2 and -9 i.e. -11 and -9. Ans.

More Examples of Word Problems

The following Links take you to More Examples of Math Word Problems.

The product of two cosecutive integers is 812. Find them.

Thedifference of two numbers is 6 and their product is 391. Find them.

Twice the square of a number exceeds 4 times the number by 240. Find it.

For Answers See at the bottom of the Page.

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