Question

root5x^2 + 2x -3 root 5

Answer 1

Answer:

(x + √5)(√5x - 3)

Step-by-step explanation:

Given that ;

$\tt \sqrt{5} x {}^{2} + 2x - 3 \sqrt{5}$

We can factorise the given equation by splitting the middle term. We need to find two numbers such that their product is (-3√5 * √5) = (-15). And their sum is 2.

Therefore, such two numbers can be 5 and 3.

$\implies \tt \sqrt{5} x {}^{2} + 2x - 3 \sqrt{5} \\ \\ \implies \tt \sqrt{5} {x}^{2} + 5x - 3x - 3 \sqrt{5} \\ \\ \implies \tt \sqrt{5} x( x + \sqrt{5} ) - 3(x + \sqrt{5} ) \\ \\ \implies \tt (x + \sqrt{5} )( \sqrt{5} x - 3)$

Hence, the answer is (x + √5)(√5x - 3).

Answer 2

Answer:

$\huge{\boxed{\sf{x=-\sqrt{5}\:and\frac{3}{\sqrt{5}}}}}$

Step-by-step explanation:

$\huge{\boxed{\sf{SOLUTION-}}}$

$\huge{\boxed{\sf{METHOD(1)-}}}$

$\sqrt{5} {x}^{2} + 2x - 3 \sqrt{5} = 0 \\ \sqrt{5} {x}^{2} + 5x - 3x - 3 \sqrt{5} = 0 \\ \sqrt{5} x(x + \sqrt{5} ) - 3(x + \sqrt{5} ) = 0 \\ (\sqrt{5} x - 3)(x + \sqrt{5} ) = 0 \\ \sqrt{5} x - 3 = 0 \\ \sqrt{5} x = 3 \\ x = \frac{3}{ \sqrt{5} } - - - - - 1st \: root \\ x + \sqrt{5} = 0 \\ x = - \sqrt{5} - - - - - 2nd \: root$

$\huge{\boxed{\sf{METHOD(2)-}}}$

$d = {b}^{2} - 4ac \\ d = 4 + 60 \\ d = 64 \\ x = \frac{ - b + \sqrt{d} }{2a} \\ x = \frac{ - 2 + 8}{2 \times \sqrt{5} } \\ x = \frac{3}{ \sqrt{5} } \\ x = \frac{3 \sqrt{5} }{5} - - - - 1st \: root \\ x = \frac{ - 2 - 8}{ \sqrt{5} } \\ x = \frac{ - 10}{2 \sqrt{5} } \\ x = -\sqrt{5} - - - - - 2nd \: root$

$\huge{\boxed{\sf{x=-\sqrt{5}\:and\frac{3}{\sqrt{5}}}}}$