Question

FACTORIES
$5 \sqrt{5 } {x}^{2} + 20x + 3 \sqrt{5}$
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Answer 1

Factorize : $5 \sqrt{5 } {x}^{2} + 20x + 3 \sqrt{5} = 0$

Comparing the given equation with ax^2 + bx + c = 0, we get =>

a = 5√5

b = 20

c = 3√5

Now,

$\tt {b^2 - 4ac = (20) {}^{2} - 4 \times 5 \sqrt{5} \times 3 \sqrt{5}}$

$\tt {= 400 - 4 \times 15 \times 5}$

$\tt {= 400 - 300}$

$= 100$

b^2 - 4ac = 100.

Now, Using Formula Method,

$\tt{x = \frac{ - b (+ - ) \sqrt{b {}^{2} - 4ac } }{2a}}$

$\tt{ = \frac{ - 20( + - ) \sqrt{100} }{2 \times 5 \sqrt{5}} }$

$\tt{= \frac{ - 20( + - )10}{10 \sqrt{5} }}$

$\tt { = \frac{-20 + 10}{10 \sqrt{5} } \: \: \: \: or \: \: \: \: \frac{-20 - 10}{10 \sqrt{5} }}$

$\tt{ = \frac{-30}{10 \sqrt{5} } \: \: \: or \: \: \: \frac{-10}{3 \sqrt{5}} }$

$\tt {= \frac{-3}{ \sqrt{5} } \: \: \: or \: \: \: \frac{-1}{ \sqrt{5}} }$

Hence,

(X = -3/√5) or (X = -1/√5)

Answer 2

$\textbf{\huge{ANSWER:}}$

$\sf{Given\:equation:}$

$5\sqrt{5}x^{2} + 20x + 3\sqrt{5}$

Let's do this by the splitting middle term method :-

$5\sqrt{5}x^{2} + 20x + 3\sqrt{5}$

=》 $5\sqrt{5}x^{2} + 15x + 5x + 3\sqrt{5}$

=》 $5x ( \sqrt{5} + 3) + \sqrt{5}(\sqrt{5} x + 3)$

=》 $(5x + \sqrt{5}) (\sqrt{5} x + 3)$

Now, we've simplified the L.H.S.

Let's now equal L.H.S. and R.H.S.

Start with keeping the values equal to zero, as the R.H.S. = 0 :-

=》 $5x + \sqrt{5} = 0$

=》 $\boxed{x = \frac{(-1)}{\sqrt{5}}}$

Or

=》 $\boxed{x = \frac{(-3)}{\sqrt{5}}}$

The value of x can vary between both the values obtained by solving. You can again go through the answer and ask your doubts in the comments section.