Answer:
x= [tex]-2 \pm \sqrt{13}[/tex]
Step-by-step explanation:
Given: [tex]3x^{2} -12x+27= 0[/tex]
This is a quadratic equation.
[tex]3x^{2} -12x+27= 0[/tex]
Taking common 3 from the equation given
⇒ [tex]3(x^{2} -4x+9)=0[/tex]
dividing both side by 3
⇒ [tex]x^{2} -4x+9=0[/tex]
By using different factor, we can observe this trinomial can not be factored.
∴ solving it by completing the square.
⇒ [tex]x^{2} -4x+9=0[/tex]
adding 9 on both side
⇒ [tex]x^{2} -4x= 9[/tex]
To get a perfect square number as coeffecient of x is 4, lets add both side by 4.
⇒[tex]x^{2} -4x+4= 13[/tex]
we know, [tex](x+2)^{2} = x^{2} +2\times x\times 2+2^{2}[/tex]
∴ [tex](x+2)^{2} =13[/tex]
Taking Square root on both side, remember; √a²= a
⇒ [tex](x+2)= \sqrt{13}[/tex]
opening parenthesis
⇒ [tex]x+2= \sqrt{13}[/tex]
subtracting both side by 2
∴ x= [tex]-2 \pm \sqrt{13}[/tex]
