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 :
- https://brainly.com/question/10449635 [Answered by Calculista]
- https://brainly.com/question/1596209 [Answered by AkhileshT]
Keywords :
Quadratic equation, standard form, completing squares method, Polynomial identities.
