Question

find the roots of 3X^2+11X+10=0 by completing square method​

Answer 1

Step-by-step explanation:

3x^2+11x+10=0

transpose 10to RHS

=> 3x^2+11x=-10

make coefficient of x^2 =1

( divide whole by 3)

$\frac{ {3x}^{2} }{3} + \frac{11x}{3} = \frac{ - 10}{3}$

=>

${x}^{2} + \frac{11}{3} x = \frac{ - 10}{3}$

take

$\frac{11}{3}$

and make it half and square it.....

=>

$\frac{11}{2} \times \frac{1}{2}$

$( \frac{11}{6} )$

square it and add it on both sides

${x}^{2} + \frac{11}{3} x + { \frac{11}{6} }^{2} = \frac{ - 10}{3} + { \frac{11}{6} }^{2}$

simplify LHS

$( {x + \frac{11}{6} }^{2} ) = \frac{ - 10}{3} + \frac{121}{36}$

$( {x + \frac{11}{6} }^{2} ) = \frac{1}{36}$

$\sqrt{( {x + \frac{11}{6} )}^{2} } = + or - \sqrt{ \frac{1}{36} }$

$x + \frac{11}{6} = \frac{1}{6}$

$x = \frac{1 - 11}{6} = > \: \: \: x = \frac{ - 10}{6} = > \: \: x = \frac{ - 5}{3} and \: x = \frac{ - 1 - 11}{6} = > \frac{ - 12}{6} = > x = - 2$

$= > \: \: x = \frac{ - 5}{3} or \: x = - 2$

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