Math, asked by hamzakhan1207, 1 year ago

Equation  25x^2-2xy+\frac{1}{25}y^2=(5x+ky)^2 is true when k is :
(A)  \frac{1}{5}
(B)  -\frac{1}{5}
(C) 1
(D) − 1

Answers

Answered by TooFree
1

Answer:

(B) -1/5


Step-by-step explanation:

25x^2-2xy+\dfrac{1}{25}y^2=(5x+ky)^2


RHS:

(5x+ky)^2 = (5x)^2 + 2(5x)(ky) + (ky)^2

(5x+ky)^2 = 25x^2 + 10kxy + k^2y^2


Comparing the LHS and RHS:

25x^2-2xy+\dfrac{1}{25}y^2 = 25x^2 + 10kxy + k^2y^2

-2xy = 10kxy

-2 = 10k

k = -2 \div 10

k = -\dfrac{1}{5}


Answer: (B) -1/5

Answered by rohitkumargupta
1

HELLO DEAR,




GIVEN:-

25x² - 2xy + y²/25 = (5x + ky)²


So , 25x² - 2xy + y²/25 = 25x² + k²y² + 10kxy



On comparing BOTH side,


We get,


-2xy = 10kxy & y²/25 = k²y²


=> K = -2/10 & k² = 1/25


=> K = -1/5 & k = ±1/5


Hence, option (b) is correct [as -1/5 satisfied both]



I HOPE IT'S HELP YOU DEAR,

THANKS

Similar questions