Question

Solve by completing the square
3x^2-9x -7 = -3x

Answer 1

First, let's simplify the equation by moving the term -3x from the right side to the left side:

`3x^2 - 9x + 3x - 7 = 0`

This simplifies to:

`3x^2 - 6x - 7 = 0`

Next, we divide all terms by 3 to make the coefficient of

`x^2`

equal to 1:

`x^2 - 2x - (7)/(3) = 0`

Now, we complete the square. To do this, we take half of the coefficient of x, square it, and add it to both sides of the equation. Half of -2 is -1, and -1 squared is 1. So we add 1 to both sides:

`x^2 - 2x + 1 = (7)/(3) + 1`

This gives us:

`x^2 - 2x + 1 = (10)/(3)`

The left side of the equation can now be written as a perfect square:

`(x - 1)^2 = (10)/(3)`

Finally, we solve for x by taking the square root of both sides:

`x - 1 = \pm \sqrt{(10)/(3)}`

So the solutions are:

`x = 1 \pm \sqrt{(10)/(3)}`

Answer 2

`x = \sqrt{{(10)/(3)}}+1, \ -\sqrt{{(10)/(3)}}+1`

Solved Stepwise:

`\sf \rightarrow 3x^2-9x -7 = -3x`

`\sf \rightarrow 3x^2-9x +3x-7 =0`

`\sf \rightarrow 3x^2-6x-7 =0`

`\sf \rightarrow 3(x^2 - 2x)-7 =0`

`\sf \rightarrow 3(x - (2)/(2) )^2-7-3((-2)/(2))^2 =0`

`\sf \rightarrow 3(x -1)^2-7-3(-1)^2 =0`

`\sf \rightarrow 3(x -1)^2-7-3 =0`

`\sf \rightarrow 3(x -1)^2-10 =0`

`\sf \rightarrow 3(x -1)^2 = 10`

`\sf \rightarrow (x -1)^2 = (10)/(3)`

`\sf \rightarrow x -1 = \pm \sqrt{{(10)/(3)}`

`\sf \rightarrow x = \pm \sqrt{{(10)/(3)}}+1`

`\sf \rightarrow x = \sqrt{{(10)/(3)}}+1, \ -\sqrt{{(10)/(3)}}+1`