Question

use suitable identities and find the following products: (III) (√2x+√3y) (√2x+√3y) ye under root ka sign hai Jo 2x ke aage LGA h please give me my ANS fast ​

Answer 1

Answer:

$( \sqrt{2} x + \sqrt{3} y)( \sqrt{2} x + \sqrt{3} y) \\ \\ = ( \sqrt{2} x + \sqrt{3} y) ^{2} \\ \\ = { (\sqrt{2}x) }^{2} + 2 \times \sqrt{2} x \times \sqrt{3} y + ( { \sqrt{3} y})^{2} \\ \\ = 2x ^{2} + 2 \sqrt{6} xy + 3 {y}^{2} \\ \\ formula \: used \\ \\ (a + b) ^{2} = a ^{2} + 2ab + b ^{2}$

hope it helps uH ✔

Answer 2

Step-by-step explanation:

The required solution to the given system is

x = 0, y = 0.

Step-by-step explanation: We are given to solve the following system of equations by the method of cross-multiplication :

\begin{gathered}\sqrt2x+\sqrt3y=0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)\\\\\sqrt3x-\sqrt8y=0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(ii)\end{gathered}

2

x+

3

y=0 (i)

3

x−

8

y=0 (ii)

The table of cross-multiplication is as follows :

x y 1

√3 0 √2 √3

-√8 0 √3 -√8

Therefore, by the method of cross-multiplication, we get

\begin{gathered}\dfrac{x}{\sqrt3\times0-0\times(-\sqrt8)}=\dfrac{y}{0\times\sqrt3-\sqrt2\times0}=\dfrac{1}{\sqrt2\times(-\sqrt8)-\sqrt3\times\sqrt3}\\\\\\\Rightarrow \dfrac{x}{0}=\dfrac{y}{0}=\dfrac{1}{-4-3}\\\\\\\Rightarrow x=\dfrac{0}{-7}=0,~~y=\dfrac{0}{-7}=0.\end{gathered}

3

×0−0×(−

8

)

x

=

0×

3

−

2

×0

y

=

2

×(−

8

)−

3

×

3

1

⇒

0

x

=

0

y

=

−4−3

1

⇒x=

−7

0

=0, y=

−7

0

=0.

Thus, the required solution to the given system is

x = 0, y = 0.