Question

Complete the square: ax^2 + x + 3.

Answer 1

Answer:

The value of a is $\frac{1}{12}$.

Step-by-step explanation:

The given expression is

$ax^2+x+3$

A quadratic expression $ax^2+bx+c$ is complete square if $b^2-4ac=0$

For the given expression a=a,b=1 and c=3.

$(1)^2-4(a)(3)=0$

$1-12a=0$

Add 12a on both sides.

$1=12a$

Divide both sides by 12.

$\dfrac{1}{12}=a$

Therefore, the value of a is $\frac{1}{12}$.

$\frac{1}{12}x^2+x+3$

$(\frac{1}{2\sqrt{3}}x)^2+2(\frac{1}{2\sqrt{3}}x)(\sqrt{3}+(\sqrt{3})^2$

$(\frac{1}{2\sqrt{3}}x+\sqrt{3})^2$           $[\because (a+b)^2=a^2+2ab+b^2]$

Answer 2

Answer:

$a(x+\frac{1}{2a})^2)-\frac{1}{4a}+3$

Step-by-step explanation:

We have been given an expression $ax^2+x+3$. We are asked to complete the square for the given expression.

First of all, we will factor our a as:

$a(x^2+\frac{x}{a})+3$  

$a(x^2+\frac{1}{a}x)+3$

Now, we need to add and subtract half the square of the middle term, that is $(\frac{1}{2a})^2$:

$a[x^2+\frac{1}{a}x+(\frac{1}{2a})^2-(\frac{1}{2a})^2]+3$      

$a[x^2+\frac{1}{a}x+(\frac{1}{2a})^2-(\frac{1}{4a^2})]+3$    

$a(x^2+\frac{1}{a}x+(\frac{1}{2a})^2)+3-a*\frac{1}{4a^2}$  

$a(x+\frac{1}{2a})^2)-\frac{1}{4a}+3$

Therefore, our required square would be $a(x+\frac{1}{2a})^2)-\frac{1}{4a}+3$.