Question

Solve the equation....???
$6 {x }^{2} + 3x + 1$
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Answer 1

$\sf\huge {\underline{\underline{{Solution}}}}$

$\sf\implies6 {x}^{2} + 3x + 1$

We have to solve the Equation 6x²+3x+1

We can solve it by factorising it through quadratic formula:

Quadratic formula:

$\sf \boxed{ \: x = \dfrac{ - b \pm \sqrt{ {b}^{2} - 4ac} }{2a} }$

Finding the solution using quadratic formula:

a= 6;b = 3 & c= 1

$\sf\implies \: x = \dfrac{ - 3 \pm \sqrt{ {(3)}^{2} - 4(6)(1)} }{2( 6)}$

$\sf\implies \: x = \dfrac{ - 3 \pm \sqrt{9 - 24} }{12}$

$\sf\implies \: x = \dfrac{ - 3 \pm \sqrt{ - 15} }{12}$

Here,the discriminant b²-4ac is <0

So,there will be two complex roots of the Equation.

$\implies \: x = \dfrac{ - 3 \pm \sqrt{15i} }{12}$

$\implies \: x = - \dfrac{3}{12} \pm \dfrac{ \sqrt{15i} }{12}$

$\implies \: x = - \dfrac{1}{4} \pm \dfrac{ \sqrt{15i} }{12}$

$x = - 0.25 + 0.322749i$

$x = 0.25 - 0.322749i$

Note : the symbol $i$(iota) is used where there is negative value inside root.It is used in place of negative sign.