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Sagot :
For this case we have the following polynomial:
[tex]3x^2-2x+7=0[/tex]
To solve the problem, we must complete squares.
The first step is to divide the entire expression by 3.
We have then:
[tex]\frac{3}{3}x^2-\frac{2}{3}x+\frac{7}{3}=0[/tex]
The second step is to place the constant term on the right side of the equation:
[tex] \frac{3}{3}x^2-\frac{2}{3}x=-\frac{7}{3} [/tex]
The third step is to complete the square:
[tex] \frac{3}{3}x^2-\frac{2}{3}x + (-\frac{1}{3})^2=-\frac{7}{3}+ (-\frac{1}{3})^2 [/tex]
Rewriting we have:
[tex] x^2-\frac{2}{3}x + \frac{1}{9}=-\frac{7}{3}+ \frac{1}{9} [/tex]
[tex] (x-\frac{1}{3})^2 = -\frac{20}{3} [/tex]
Answer:
By completing squares we have:
[tex] (x-\frac{1}{3})^2 = -\frac{20}{3} [/tex]
Answer:
[tex]-\dfrac{20}{9}[/tex]
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 : [tex]3x^2-2x+7=0[/tex]
First we divide both sides by 3 , we get
[tex]x^2-\dfrac{2}{3}x+\dfrac{7}{3}=0[/tex]--------(1)
Compare this equation to [tex]x^2+2ax+a^2[/tex] , we have
[tex]2a=\dfrac{-2}{3}[/tex]
[tex]\Rightarrow\ a=\dfrac{-1}{3}[/tex] [divide both sides by 2]
Now using the completing the squares method , Add and subtract [tex](\dfrac{-1}{3})^2[/tex] to the left side in (1), we get
[tex]x^2-\dfrac{2}{3}x+(\dfrac{-1}{3})^2-(\dfrac{-1}{3})^2+\dfrac{7}{3}=0[/tex]
It can be written as
[tex](x^2-2(\dfrac{1}{3})x+\dfrac{1}{3})^2)-\dfrac{1}{9}+\dfrac{7}{3}=0[/tex]
Use identity [tex]x^2-2ax+a^2=(x-a)^2[/tex], we have
[tex](x-\dfrac{1}{3})^2)+\dfrac{7(3)-1}{9}=0[/tex]
[tex](x-\dfrac{1}{3})^2)+\dfrac{20}{9}=0[/tex]
Subtract [tex]\dfrac{20}{9}[/tex] from both the sides , we get
[tex](x-\dfrac{1}{3})^2)=-\dfrac{20}{9}[/tex]
Therefore, the value of [tex](x-\dfrac{1}{3})^2)=-\dfrac{20}{9}[/tex]
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.
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