If 3x^2-2x+7=0 then (x-1/3)^2

Asked by maham237 @ 25/12/2020 in Mathematics viewed by 395 People

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

Answered by maham237 @ 25/12/2020

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.

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