Question

please don't give irritating answer solve fast it's urgent​

Answer 1

AnsWer :

y = √3 and y = -3√3.

Solution :

We have Equation,

=> y² +2√3y -9 = 0.

=> y² + 3√3y -√3 -9 = 0.

=> y(y+ 3√3) -√3 (y + 3√3 ) = 0.

=> (y-√3)(y + 3√3).

$\rule{40}1$

Either,

=> y + 3√3 = 0.

=> y = -3√3.

$\rule{40}1$

Or,

=> y-√3 = 0.

=> y = √3.

$\rule{200}3$

Now, Let try by Quadratic Formula,

$\tt \dagger \: \: \: x = \frac{ - b \pm \sqrt{ {b}^{2} - 4ac } }{2a}$

we have,

• a = 1.
• b = 2√3.
• c = -9.

Putting all given value.

$\tt\longmapsto \frac{ - 2 \sqrt{3} \pm \sqrt{ ({2 \sqrt{3} )}^{2} - 4 \times 1 \times - 9 } }{2}$

$\tt\longmapsto \frac{ - 2 \sqrt{3} \pm \sqrt{48} }{2}$

$\tt\longmapsto \frac{ - 2 \sqrt{3} \pm4 \sqrt{3} }{2}$

$\tt\longmapsto \frac{2( - \sqrt{3} \pm2 \sqrt{3} )}{2}$

$\tt\longmapsto \frac{ - \sqrt{3} \pm2 \sqrt{3} }{2}$

$\rule{40}1$

Either,

$\tt\longmapsto y = {- \sqrt{3} - 2 \sqrt{3} }$

$\tt\longmapsto y = - 3 \sqrt{3} .$

$\rule{40}1$

Or,

$\tt\longmapsto y = - \sqrt{3} + 2 \sqrt{3} .$

$\tt\longmapsto y = \sqrt{3} .$

Therefore, the value of y be √3 and -3√3.

Answer 2

Answer:

Hey mate plzz refer the above attachment for the answer......

HOPE IT WILL HELP YOU ❤️❤️❤️

plzzz mark me as BRAINLIEST ❤️❤️❤️