Question

If 3x^2-2x+7=0,then (x-1/3)^2= please help with detailed steps because i dont really understand it. I know the answer is -20/9 but please explain

Answer 1

Answer:

$-\dfrac{20}{9}$

Explanation:

A quadratic function is a kind of function with highest degree 2 .  Standard form of the quadratic equation : tex]ax^2+bx+c=0[/tex]

Further explanation:

Consider the given quadratic equation : $3x^2-2x+7=0$  

First we divide both sides  by 3 , we get

$x^2-\dfrac{2}{3}x+\dfrac{7}{3}=0$--------(1)

Compare this equation to $x^2+2ax+a^2$ , we have

$2a=\dfrac{-2}{3}$  

$\Rightarrow\ a=\dfrac{-1}{3}$  [divide both sides by 2]

Now using the completing the squares method , Add and subtract $(\dfrac{-1}{3})^2$ to the left side in (1), we get

$x^2-\dfrac{2}{3}x+(\dfrac{-1}{3})^2-(\dfrac{-1}{3})^2+\dfrac{7}{3}=0$  

It can be written as

$(x^2-2(\dfrac{1}{3})x+\dfrac{1}{3})^2)-\dfrac{1}{9}+\dfrac{7}{3}=0$  

Use identity $x^2-2ax+a^2=(x-a)^2$, we have

$(x-\dfrac{1}{3})^2)+\dfrac{7(3)-1}{9}=0$  

$(x-\dfrac{1}{3})^2)+\dfrac{20}{9}=0$  

Subtract $\dfrac{20}{9}$ from both the sides , we get

$(x-\dfrac{1}{3})^2)=-\dfrac{20}{9}$

Therefore, the value of $(x-\dfrac{1}{3})^2)=-\dfrac{20}{9}$

Learn more :

• brainly.com/question/10449635  [Answered by Calculista]

• brainly.com/question/1596209  [Answered by AkhileshT]

Keywords :

Quadratic equation, standard form, completing squares method, Polynomial identities.

Answer 2

For this case we have the following polynomial:

$3x^2-2x+7=0$

To solve the problem, we must complete squares.

The first step is to divide the entire expression by 3.

We have then:

$\frac{3}{3}x^2-\frac{2}{3}x+\frac{7}{3}=0$

The second step is to place the constant term on the right side of the equation:

$\frac{3}{3}x^2-\frac{2}{3}x=-\frac{7}{3}$

The third step is to complete the square:

$\frac{3}{3}x^2-\frac{2}{3}x + (-\frac{1}{3})^2=-\frac{7}{3}+ (-\frac{1}{3})^2$

Rewriting we have:

$x^2-\frac{2}{3}x + \frac{1}{9}=-\frac{7}{3}+ \frac{1}{9}$

$(x-\frac{1}{3})^2 = -\frac{20}{3}$

Answer:

By completing squares we have:

$(x-\frac{1}{3})^2 = -\frac{20}{3}$