Different problems using algebraic identities :
Here we are going to see some different problems using identities.
Problem 1 :
If (x + a) (x + b) (x + c) = x³ - 10 x² + 45 x - 15 find a+b+c and 1/a+1/b+1/c and a² + b² + c²
(x + a)(x + b)(x + c) = x³ + (a+b+c)x² + (ab + bc + ca)x + abc
x³+(a+b+c)x²+(ab+bc+ca)x+abc = x³ - 10 x² + 45 x - 15
By comparing these two equations coefficients of x², x and constant are equal.
Coefficient of x² :
a + b + c = -10
Coefficient of x :
ab + bc + ca = 45
Constant term :
abc = -15
(i) a + b +c = -10
(ii)(1/a) + (1/b) + (1/c)
To find the value of (1/a) + (1/b) + (1/c), we have to find the least common multiple of denominators.
(iii) a² + b² + c²
To find the value of a² + b² + c², we have to use the algebraic identity of (a + b + c)².
(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca
(a + b + c)² = a² + b² + c² + 2(ab + bc + ca)
a² + b² + c² = (a + b + c)² - 2(ab + bc + ca)
= (-10)² - 2 (45)
= 100 - 90
Problem 2 :
Find 8x³ + 27y³ if 2x + 3y = 13 and xy = 6
8x³ + 27y³ = 2³x³ + 3³y³ ==> (2x)³ + (3y)³
a³ + b³ = (a+b)³ - 3ab (a+b)
a = 2x and b = 3y
(2x)³ + (3y)³ = (2x + 3y)³ - 3(2x)(3y)(2x + 3y)
= 13³- 18xy (13)
= 2197- 18(6) (13)
= 2197- 1404
After having gone through the stuff given above, we hope that the students would have understood "Different problems using algebraic identities".
Apart from the stuff given above, if you want to know more about "Different problems using algebraic identities", please click here
Apart from the stuff given in this section, if you need any other stuff in math, please use our google custom search here.
APTITUDE TESTS ONLINE
ACT MATH ONLINE TEST
TRANSFORMATIONS OF FUNCTIONS
ORDER OF OPERATIONS
MATH FOR KIDS
HCF and LCM word problems
Word problems on quadratic equations
Word problems on comparing rates
Converting repeating decimals in to fractions