Respuesta :
Answer:
The value of a is [tex]\frac{1}{12}[/tex].
Step-by-step explanation:
The given expression is
[tex]ax^2+x+3[/tex]
A quadratic expression [tex]ax^2+bx+c[/tex] is complete square if [tex]b^2-4ac=0[/tex]
For the given expression a=a,b=1 and c=3.
[tex](1)^2-4(a)(3)=0[/tex]
[tex]1-12a=0[/tex]
Add 12a on both sides.
[tex]1=12a[/tex]
Divide both sides by 12.
[tex]\dfrac{1}{12}=a[/tex]
Therefore, the value of a is [tex]\frac{1}{12}[/tex].
[tex]\frac{1}{12}x^2+x+3[/tex]
[tex](\frac{1}{2\sqrt{3}}x)^2+2(\frac{1}{2\sqrt{3}}x)(\sqrt{3}+(\sqrt{3})^2[/tex]
[tex](\frac{1}{2\sqrt{3}}x+\sqrt{3})^2[/tex] [tex][\because (a+b)^2=a^2+2ab+b^2][/tex]
Answer:
[tex]a(x+\frac{1}{2a})^2)-\frac{1}{4a}+3[/tex]
Step-by-step explanation:
We have been given an expression [tex]ax^2+x+3[/tex]. We are asked to complete the square for the given expression.
First of all, we will factor our a as:
[tex]a(x^2+\frac{x}{a})+3[/tex]
[tex]a(x^2+\frac{1}{a}x)+3[/tex]
Now, we need to add and subtract half the square of the middle term, that is [tex](\frac{1}{2a})^2[/tex]:
[tex]a[x^2+\frac{1}{a}x+(\frac{1}{2a})^2-(\frac{1}{2a})^2]+3[/tex]
[tex]a[x^2+\frac{1}{a}x+(\frac{1}{2a})^2-(\frac{1}{4a^2})]+3[/tex]
[tex]a(x^2+\frac{1}{a}x+(\frac{1}{2a})^2)+3-a*\frac{1}{4a^2}[/tex]
[tex]a(x+\frac{1}{2a})^2)-\frac{1}{4a}+3[/tex]
Therefore, our required square would be [tex]a(x+\frac{1}{2a})^2)-\frac{1}{4a}+3[/tex].
