Question

Solve by taking the square root I need help understanding steps this is hard for me. ​

Answer 1

Answer:

x= $-2 \pm \sqrt{13}$

Step-by-step explanation:

Given: $3x^{2} -12x+27= 0$

This is a quadratic equation.

$3x^{2} -12x+27= 0$

Taking common 3 from the equation given

⇒ $3(x^{2} -4x+9)=0$

dividing both side by 3

⇒ $x^{2} -4x+9=0$

By using different factor, we can observe this trinomial can not be factored.

∴ solving it by completing the square.

⇒ $x^{2} -4x+9=0$

adding 9 on both side

⇒ $x^{2} -4x= 9$

To get a perfect square number as coeffecient of x is 4, lets add both side by 4.

⇒$x^{2} -4x+4= 13$

we know, $(x+2)^{2} = x^{2} +2\times x\times 2+2^{2}$

∴ $(x+2)^{2} =13$

Taking Square root on both side, remember; √a²= a

⇒ $(x+2)= \sqrt{13}$

opening parenthesis

⇒ $x+2= \sqrt{13}$

subtracting both side by 2

∴ x= $-2 \pm \sqrt{13}$