Question

Math ke 10 formulae (abc)/2

Answer 1

Answer:

(a+b)2 = a2 + 2ab + b2

a2 + b2 = (a + b)2 – 2ab

(a – b)2 = a2 – 2ab + b2

(a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca

(a – b – c)2 = a2 + b2 + c2 – 2ab + 2bc – 2ca

(a + b)3 = a3 + 3a2b + 3ab2 + b3 ; (a + b)3 = a3 + b3 + 3ab(a + b)

(a – b)3 = a3 – 3a2b + 3ab2 – b3

a3 – b3 = (a – b)(a2 + ab + b2)

a3 + b3 = (a + b)(a2 – ab + b2)

(a + b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4

(a – b)4 = a4 – 4a3b + 6a2b2 – 4ab3 + b4

a4 – b4 = (a – b)(a + b)(a2 + b2)

a5 – b5 = (a – b)(a4 + a3b + a2b2 + ab3 + b4)

If n is a natural number an – bn = (a – b)(an-1 + an-2b+…+ bn-2a + bn-1)

If n is even (n = 2k), an + bn = (a – b)(an-1 + an-2b +…+ bn-2a + bn-1)

If n is odd (n = 2k + 1), an + bn = (a + b)(an-1 – an-2b +an-3b2…- bn-2a + bn-1)

(a + b + c + …)2 = a2 + b2 + c2 + … + 2(ab + ac + bc + ….)

Laws of Exponents (am)(an) = am+n ; (ab)m = ambm ; (am)n = amn

Fractional Exponents a0 = 1 ; aman=am−n ; am = 1a−m ; a−m = 1am

Roots of Quadratic Equation

For a quadratic equation ax2 + bx + c where a ≠ 0, the roots will be given by the equation as −b±b2−4ac√2a

Δ = b2 − 4ac is called the discriminant

For real and distinct roots, Δ > 0

For real and coincident roots, Δ = 0

For non-real roots, Δ < 0

If α and β are the two roots of the equation ax2 + bx + c then, α + β = (-b / a) and α × β = (c / a).

If the roots of a quadratic equation are α and β, the equation will be (x − α)(x − β) = 0

Factorials

n! = (1).(2).(3)…..(n − 1).n

n! = n(n − 1)! = n(n − 1)(n − 2)! = ….

0! = 1

(a+b)n=an+nan−1b+n(n−1)2!an−2b2+n(n−1)(n−2)3!an−3b3+….+bn,where,n>1

Solved Examples

Question 1: Find out the value of 52 – 32

Solution:

Using the formula a2 – b2 = (a – b)(a + b)

where a = 5 and b = 3

(a – b)(a + b)

= (5 – 3)(5 + 3)

= 2 × 8

= 16

Question 2: 43 × 42 = ?

Solution:

Using the exponential formula (am)(an) = am+n

where a = 4

43 × 42

= 43+2

= 45

= 1024