Question

Find the value using suitable identity

$101 \times 98$
​

Answer 1

9898

Step-by-step explanation:

101 x 98

( x + a ) ( x + b ) = x² + ( a + b ) x + ab

x= 100 , a= 1 ,b= -2

(100)²+ [ 1 + ( - 2 ) ] × 100 + 1  ( - 2 )

10000-100-2

9900-2

9898

Answer 2

$\Large{\underbrace{\sf{\orange{Required\:Solution:}}}}$

101×98

Solution - $\displaystyle{\sf{ (100+1)(100-2)}}$

• Using Identity - (x+α)(x+b) = x²+(α+b)x+ αb

$\longrightarrow{\sf{ (100^{2} + \big[1+(-2)\big]\times 100 + \big[(1 \times(-2)\big]}}$

$\longrightarrow{\sf{10,000 + (-1)\times100 + (-2)}}$

$\longrightarrow{\sf{ 10,000 + (-100) +(-2)}}$

$\longrightarrow{\sf{ 10,000 + (-102)}}$

$\large\implies{\boxed{\sf{\orange{9,898}}}}$

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$\Large{\underbrace{\sf{\orange{Explore\:More!}}}}$

Algebrαic identity - An αlgebrαic identity is αn equαlity thαt holds for αny vαlues of its vαriαbles.

There αre some αlgebrαic identities —

• (a+b)² = a² + 2ab + b²
• (a-b)² = a² - 2ab + b²
• a²-b² = (a-b)(a+b)
• a²+b² = (a+b)² - 2ab
• (x+a)(x+b) = x² + (a+b)x + ab
• (a+b)³ = a³ + b³ + 3ab(a+b)
• (a-b)³ = a³ - b³ - 3ab(a-b)
• a³+b³ = (a+b)³-3ab(a+b)

⠀⠀⠀⠀⠀⠀⠀= (a+b) (a²- ab + b²)

• a³-b³ = (a-b)³+3ab(a-b)

⠀⠀⠀⠀⠀⠀⠀= (a-b) (a² + ab + b²)

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

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