Question

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

Answer 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

Answer 2

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