USING ALGEBRAIC IDENTITES PRACTICE PROBLEMS

Problem 1 :

If x-y  =  2 and xy  =  24, what is the value of x + y ?

Problem 2 :

If

a+b+c  =  15 and a²+b²+c²  =  83

what is the value of ab+bc+ca ?

Problem 3 :

If a-b  =  4 and ab  =  60, what is the value of a + b ?

Problem 4 :

If x-1/x  =  4, prove that x⁴ + 1/x⁴  =  322

Problem 5 :

If x - y  =  8 and xy  =  5, what is the value of

x³-y³ + 8(x+y)

Problem 6 :

If a - b  =  5 and ab  =  36, find the value of

(i)  a²+ab+b²

(ii)  a³-b³

Problem 7 :

If x + y  =  5, xy  =  6 and x > y

(i)  Find the value of 2(x² + y²)

(ii)  Find the value of x³-y³-3(x²+y²)

(iii)  Find the value of x⁵+y⁵.

(1)  Solution :

(x + y)²  =  x²+2xy+y²   ----(1)

(x - y)²  =  x²-2xy+y² ----(2)

From (2)

x²+y²  =  (x - y)² + 2xy

By applying the value of x²+y² in (1), we get

(x + y)²  =  2xy + (x - y)² + 2xy

(x + y)²  =  (x - y)² + 4xy

Here x - y  =  2 and xy  =  24

(x + y)²  =  (2)² + 4(24)

(x + y)²  =  4 + 96

(x + y)  =  √100

x+y  =  ±10

(2)  Solution :

(a+b+c)²  =  a²+b²+c²+2ab+2bc+2ca

By applying the given value, we get

(a+b+c)²  =  a²+b²+c²+2(ab+bc+ca)

15²  =  83+2(ab+bc+ca)

225-83  =  2(ab+bc+ca)

142  =  2(ab+bc+ca)

ab+bc+ca  =  71

(3)  Solution :

(a + b)²  =  a²+2ab+b²   ----(1)

(a - b)²  =  a²-2ab+b² ----(2)

From (2)

a²+b²  =  (a - b)² + 2ab

By applying the value of a²+b² in (1), we get

(a + b)²  =  2ab + (a - b)² + 2ab

(a + b)²  =  (a - b)² + 4ab

Here a - b  =  4 and ab  =  60

(a + b)²  =  4² + 4(60)

(a + b)²  =  16 + 240

(a + b)  =  √256

a+b  =  ±16

(4)  Solution :

Given :

x-1/x  =  4

Taking squares on both sides, we get

(x-1/x)²  =  4²

x²-2x(1/x)-1/x²  =  16

x²-2+1/x²  =  16

x²+1/x²  =  18

Again taking squares on both sides, we get

(x²+1/x²)²  =  18²

x⁴+1/x⁴ + 2  =  324

x⁴+1/x⁴  =  322

(5)  Solution :

(a-b)³  =  a³-3a²b+3ab²-b³

 a³-b³  =  (a-b)³-3ab(a-b)

x³-y³  =  (x-y)³-3xy(x-y)

x³-y³  =  8³-3(5)(8)

x³-y³  =  512-120

x³-y³  =  392

x³-y³ + 8(x+y)  =  392 + 8(8)

x³-y³ + 8(x+y)  =  392+64

x³-y³ + 8(x+y)  =  456

(6)  Solution :

(i)  a²+ab+b² 

(a-b)²  =  a²-2ab+b²

5²  =  a²-2(36)+b²

a²+b²  =  25+72

a²+b²  =  97

a²+b² + ab  =  97+36

a² + ab + b²  =  133

(ii)  a³-b³

a³-b³  =  (a-b)(a²+ab+b²)

a³-b³  =  (5)(133)

a³-b³  =  665

(7)  Solution :

(i)  2(x² + y²) :

(x+y)²  =  x²+y²+2xy

5²  =  x²+y²+2(6)

25-12  =  x²+y²

 x²+y²  =  13

2(x²+y²)  =  26

(ii) x³-y³-3(x²+y²)

x³-y³  =  (x-y)³+3xy(x-y)  --(1)

(x-y)  =  √(x+y)²-4xy

(x-y)  =  √5²-4(6)

(x-y)  =  ±1

x³-y³-3(x²+y²)  =  19-3(13)  ==>  -20

x³-y³-3(x²+y²)  =  -19-3(13)  ==>  -58

Given x+y  =  5 ----(2)

x-y  =  1  ----(3)

(2) + (3)

2x  =  6 and x  =  3

So, y  =  2

(iiii)  x⁵+y⁵  

x⁵+y⁵    =  3⁵+2⁵

=  243+32

=  275

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