Algebra is a most essential part of the 9th grade math, where it is also useful for further higher grade of math. Algebra is useful for every step of higher grade math and research oriented math even. There are some examples are given below which will definitely make you understand about initial or primary stage of algebra.
Example.1) Expand (2a + 3b)²
Ans.) (2a + 3b)²
=> (2a)² + 2 . (2a) . (3b) + (3b)²
[ Using the formula (x + y)² = x² + 2xy + y² ]
=> 4a² + 12ab + 9b² (Ans.)
Example.2) Expand (a + 2b + 3c)²
Ans.) (a + 2b + 3c)²
=> (a)² + (2b)² + (3c)² + 2 {(a . 2b) + (2b . 3c) + (a . 3c)}
[ using (x + y + z)² = x² + y² + z² + 2xy + 2yz + 2zx ]
=> a² + 4b² + 9c² + 4ab + 12 bc + 6 ca (Ans.)
1 1
Example.3) Expand ( ------ x + ------ y )²
2 3
1 1
( ------ x + ------ y )²
2 3
x x y y
=> ( ------ )² + 2. ------ . ------ + ( ------ )²
2 2 3 3
[ Using the formula (x + y)² = x² + 2xy + y² ]
x² xy y²
=> ------ + ------ + ------ (Ans.)
4 3 9
1 1
Example.4) If x + ------ = 4, find the values of ( x² + ------ )
x x²
1
Ans.) Given x + ------ = 4
x
1
=> ( x + ------ )² = 4² [ squaring both the sides ]
x
1 1
=> x² + ------ + 2 . x . ------ = 16
x² x
1
=> x² + ------ = 16 – 2 = 14
x²
1
=> x² + ------ = 14 (Ans.)
x²
1 1
Example.5) If x + ------ = 5 , find the values of ( x⁴ + ------ )
x x⁴
1
Ans.) Given x + ------ = 5
x
1
=> ( x + ------ )² = 5² [ squaring both the sides ]
x
1
=> x² + ------ + 2 = 25
x²
[ Using the formula (x + y)² = x² + 2xy + y² ]
1
=> ( x² + ------ )² = 23² [ squaring both the sides ]
x²
1 1
=> (x²)² + 2 . x². ------ + ( ------ )² = 529
x² x²
1
=> x⁴ + ------- = 529 – 2 = 527
x⁴
1
=> x⁴ + ------ = 527 (Ans.)
x⁴
1 1
Example.6) If ( x² + ------ ) = 14, find the value of x + ------ = ?
x² x
1
Ans.) Given, ( x² + ------ ) = 14
x²
1 1
=> ( x + ------ )² - 2 . x . ------- = 14
x x
[ as per formulae (a² + b²) = (a + b)² - 2ab ]
1
=> ( x + ------ )² = 14 + 2 = 16
x
1
=> ( x + ------ )² = 4²
x
1
=> x + ------ = 4
x
1 1
Example.7) If ( x² + ------ ) = 18, find the value of x - ------ = ?
x² x
1
Ans.) Given, ( x² + ------ ) = 18
x²
1 1
=> x² + ------ - 2 . x . ------ = 18 - 2
x² x
[ as per formulae (a - b)² = a² + b² - 2ab ]
1
=> ( x - ------ )² = 16
x
1
=> x - ------- = 4 (Ans.)
x
1 1
Example.8) If ( x² + ------ ) = 18, find the value of x² - ------ = ?
x² x²
1
Ans.) x² - ------
x²
1 1 1
=> x² - ------- = ( x + ------- ) ( x - ------- )
x² x x
[ as per the formulae a² - b² = (a – b) (a + b) ]
=> 4 X 4 = 16 (Ans.)
1 47 1
Example.9) If ( x² + ----- ) = -----, find the value of x - ----- = ?
25x² 5 5x
1 47
=> ( x² + ------ ) = -------
25x² 5
1 2 47 2
=> ( x² + ------ ) - ------- = ------- - -------
25x² 5 5 5
1 2 1 47 2
=> (x)² + ------ - ------ X (x) X ------ = ------- - -------
(5x)² 1 (5x) 5 5
1 45
=> (x - ------ )² = ------- = 9 = 3²
5x 5
1
=> (x - ------- ) = 3 (Ans.)
5x
a b
Example.10) If ------ = ------ , then prove that (a + b + c)
b c
(a – b + c) = (a² + b² + c²)
Ans.) a b
Let, ------- = -------- = k
b c
and, we get from the above relation a = bk, and b = ck
put these values in required identity and show that L.H.S = R.H.S
L.H.S = (a + b + c) (a – b + c)
= (bk + ck + c) (bk – ck + c)
= (ck² + ck + c)(ck² - ck + c)
= c² (k² + k + 1) (k² - k + 1)
= c² (k⁴ + kᶟ + k² - kᶟ - k² - k + k² + k + 1)
= c² (k⁴ + k² + 1)
R.H.S = (a² + b² + c²)
= (b²k² + c²k² + c²)
= (c²k⁴ + c²k² + c²)
= c² (k⁴ + k² + 1)
So, here it has proved that L.H.S = R.H.S
So, it is proved (a + b + c) (a – b + c) = (a² + b² + c²)