Question

$3 \sqrt{5 } x {}^{2} + 25x - 10 \sqrt{5} = 0 \: \: solve \: the \: following \: quadratic \: equation \: by \: factorizatin$
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Answer 1

Correct Question :

$\dagger\: \tt 3 \sqrt{5 } x {}^{2} + 25x + 10 \sqrt{5} = 0 \: \: Solve \: the \: following \: quadratic \: equation \: by \: factorization method.$

AnsWer :

• x = -10/ 3√5 and x = -√5.

Solution :

We have Quadratic equation,

$\tt\longmapsto 3 \sqrt{5} {x}^{2} + 25x + 10 \sqrt{5} = 0.$

$\tt\longmapsto 3 \sqrt{5} {x}^{2} + 15x + 10x + 10 \sqrt{5} = 0.$

$\tt\longmapsto 3 \sqrt{5} x(x + \sqrt{5} ) + 10(x + \sqrt{5} ) = 0.$

$\tt\longmapsto( 3 \sqrt{5} x + 10)(x + 15) = 0.$

Either,

$\tt\mapsto 3 \sqrt{5} x + 10 = 0.$

$\tt \mapsto x = \frac{ - 10}{3 \sqrt{5} }$

Or,

$\tt\mapsto x + \sqrt{5} = 0.$

$\tt \mapsto x = - \sqrt{5} .$

$\rule{180}3$

Let try to Solve by Quadratic Formula,

We have,

• a = 3√5.
• b = 25.
• c = 10√5.

We have Quadratic Formula,

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

Putting given value, We get.

$\tt \longmapsto x = \frac{ - 25 \pm \sqrt{ {(25)}^{2} - 4 \times 3 \sqrt{5} \times 10 \sqrt{5} } }{6 \sqrt{5} }$

$\tt\longmapsto x = \frac{ - 25 \pm \sqrt{625 - 600} }{6 \sqrt{5} }$

$\tt\longmapsto x = \frac{ - 25 \pm \sqrt{25} }{6 \sqrt{5} }$

$\tt\longmapsto x = \frac{ - 25 \pm5}{6 \sqrt{5} }$

Either,

$\tt \mapsto x = \frac{ - 25 - 5}{6 \sqrt{5} }$

$\tt\mapsto \frac{ - 30}{6 \sqrt{5} }$

$\tt\mapsto \frac{ - 5}{ \sqrt{5} }$

Rationalizing,the denominator.

$\tt\mapsto \frac{ - 5}{ \sqrt{5} } \times \frac{ \sqrt{5} }{ \sqrt{5} }$

$\tt \mapsto - \sqrt{5} .$

$\rule{50}1$

Or,

$\tt\mapsto \frac{ - 25 + 5}{6 \sqrt{5} }$

$\tt\mapsto \frac{ - 20}{6 \sqrt{5} }$

$\tt\mapsto x = \frac{ - 10}{3 \sqrt{5} }$

Therefore, the value value of x be -10/3√5 and -√5