Question

Complete the square to make a perfect square trinomial. Then, write the result as a binomial squared. q2+11q

Answer 1

Answer:

$(q+\frac{11}{2})^2-\frac{121}{4}$

Step-by-step explanation:

We have been given an expression $q^2+11q$. We are asked to complete the square to make a perfect square trinomial. Then, write the result as a binomial squared.

We know that a perfect square trinomial is in form $a^2+2ab+b^2$.

To convert our given expression into perfect square trinomial, we need to add and subtract $(\frac{b}{2})^2$ from our given expression.

We can see that value of b is 11, so we need to add and subtract $(\frac{11}{2})^2$ to our expression as:

$q^2+11q+(\frac{11}{2})^2-(\frac{11}{2})^2$

Upon comparing our expression with $(a+b)^2=a^2+2ab+b^2$, we can see that $a=q$, $2ab=11q$ and $b=\frac{11}{2}$.

Upon simplifying our expression, we will get:

$(q+\frac{11}{2})^2-\frac{11^2}{2^2}$

$(q+\frac{11}{2})^2-\frac{121}{4}$

Therefore, our perfect square would be $(q+\frac{11}{2})^2-\frac{121}{4}$.

Answer 2

Answer:

$(q+\frac{11}{2})^2-\frac{121}{4}$

Step-by-step explanation:

The given expression is

$q^2+11q$

We need to write the result as a binomial square.

If an expression is $x^2+bx$, then we need to add $(\frac{b}{2})^2$ in the expression to make it perfect square.

In the given expression b=11 and x=q.

$(\frac{b}{2})^2=(\frac{11}{2})^2$

Add and subtract $(\frac{11}{2})^2$ in the given expression.

$q^2+11q+(\frac{11}{2})^2-(\frac{11}{2})^2$

$(q+\frac{11}{2})^2-(\frac{11}{2})^2$         $[\because (a+b)^2=a^2+2ab+b^2]$

$(q+\frac{11}{2})^2-\frac{121}{4}$

Therefore, the result as a binomial square is $(q+\frac{11}{2})^2-\frac{121}{4}$.