Here we present Solved Examples and Exercise problems on application of those Formulas.

Example 1 of Solve Algebra Problems

Solve Algebra Problems given below.

Find (i) (100.5)^{2} (ii) (100.5)^{3} using Algebra Formulas

Solution to Example 1 of Solve Algebra Problems : (i) Let P = (100.5)^{2} = (100 + 0.5)^{2} This looks like (a + b)^{2} with (100) in place of a and (0.5) in place of b We have (a + b)^{2} = a^{2} + 2ab + b^{2} (See Formula 1) ∴ P = (100 + 0.5)^{2} = (100)^{2} + 2(100)(0.5) + (0.5)^{2} = 10000 + 100 + 0.25 = 100(100 + 1) +0.25 = 10100 + 0.25 = 10100.25.Ans.

(ii) Let P = (100.5)^{3} = (100 + 0.5)^{3} This looks like (a + b)^{3} with (100) in place of a and (0.5) in place of b We have (a + b)^{3} = a^{3} + 3a^{2}b + 3ab^{2} + b^{3} (See Formula 7) ∴ P = (100 + 0.5)^{3} = 100^{3} + 3(100)^{2}(0.5) + 3(100)(0.5)^{2} + (0.5)^{3} = (100)^{2} (100 + 1.5) + 75 + 0.125 = (101.5)10000 + 75.125 = 1015000 + 75.125 = 1015075.125 Ans.

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Find (2x - 3y)(2x + 3y)(4x^{2} + 9y^{2}) and verify the result for x = -1, y = 1.

Solution to Example 2 of Solve Algebra Problems : Let P = (2x - 3y)(2x + 3y)(4x^{2} + 9y^{2}) (2x - 3y)(2x + 3y) looks like (a - b)(a + b) with 2x in place of a and 3y in place of b We have (a - b)(a + b) = (a + b)(a - b) = a^{2} - b^{2} (See Formula 3) ∴ (2x - 3y)(2x + 3y) = (2x)^{2} - (3y)^{2} = (4x^{2} - 9y^{2}) Using this in P, we get P = (4x^{2} - 9y^{2})(4x^{2} + 9y^{2}) This again looks like (a - b)(a + b) with 4x^{2} in place of a and 9y^{2} in place of b So, Applying Formula 3 again, we get P = (4x^{2})^{2} - (9y^{2})^{2} = 16x^{4} - 81y^{4}. Ans.

Verifying the result for x = -1, y = 1 :

We have (2x - 3y)(2x + 3y)(4x^{2} + 9y^{2}) = (16x^{4} - 81y^{4}) For x = -1, y = 1, L.H.S. = (2x - 3y)(2x + 3y)(4x^{2} + 9y^{2}) = {2(-1) - 3(1)}{2(-1) + 3(1)}{4(-1)^{2} + 9(1)^{2}) = {-5}{1}{4 + 9} = -5 x 13 = -65 For x = -1, y = 1, R.H.S. = {16(-1)^{4} - 81(1)^{4})}= 16 - 81 = -65 ∴ L.H.S. = R.H.S. (Verified.)

Example 3 of Solve Algebra Problems

Solve Algebra Problems given below.

Find 109 x 91 using Algebra Formula

Solution to Example 3 of Solve Algebra Problems : As you can see, the given product is product of sum and difference of two numbers, to which we can apply Formula 3. Let P = 109 x 91 = (100 + 9)(100 - 9) We have (a + b)(a - b) = a^{2} - b^{2} (See Formula 3) ∴ P = (100 + 9)(100 - 9) = (100)^{2} - (9)^{2} = 10000 - 81 = 9919. Ans.

Find by what expression we should multiply (16x^{2} - 20xy + 25y^{2}) to get the result as sum of two cubes.

Solution to Example 4 of Solve Algebra Problems : Let P = (16x^{2} - 20xy + 25y^{2}) = (4x)^{2} - (4x)(5y) + (5y)^{2} This looks like (a^{2} - ab + b^{2}) with (4x) in place of a and (5y) in place of b We have(a + b)(a^{2} - ab + b^{2}) = a^{3} + b^{3} (See Formula 4) ⇒ (a^{2} - ab + b^{2}) is to be multiplied with (a + b) to get (a^{3} + b^{3}) which is the sum of two cubes. ∴ P is to be multiplied with (4x + 5y) to get the result as sum of two cubes. Ans.

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Find (i) (99.5)^{2} (ii) (99.5)^{3} using Algebra Formulas

Find (p - 4q)(p + 4q)(p^{2} + 16q^{2}) and verify the result for p = 2, q = 1.

Find (30.5) x (29.5) using Algebra Formula

Find the product (2x + 3y)(4x^{2} - 6xy + 9y^{2}) without actual multiplication

For Answers See at the bottom of the page.

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