Question

How do you factor and solve \[{x^2} + 4x + 4 = 0\]?

Answer 1

Hint:Factoring reduces the higher degree equation into its linear equation. In the above given question, we need to reduce the quadratic equation into its simplest form in such a way that addition of products of the factors of first and last term should be equal to the middle term \[ - x\].

Complete step by step answer:
\[a{x^2} + bx + c\] is a general way of writing quadratic equations where a, b and c are the numbers. In the above expression, a=1, b=4, c=4.
\[{x^2} + 4x + 4 = 0\]
First step is by multiplying the term \[{x^2}\] and the constant term 4, we get \[4{x^2}\].After this, factors of \[4{x^2}\] should be calculated in such a way that their addition should be equal to x. Factors of 4 can be 2 and 2 and their addition is given below.
where \[2x + 2x = 4x\]
So, further we write the equation by equating it with zero and splitting the middle term according to the factors.
\[
{x^2} + 4x + 4 = 0 \\
\Rightarrow {x^2} + 2x + x + 4 = 0 \]
Now, by grouping the first two and last two terms we get common factors.
\[x\left( {x + 2} \right) + 2\left( {x + 2} \right) = 0\]
Taking x common from the first group and 2 common from the second we get the above equation.
\[
\left( {x + 2} \right)\left( {x + 2} \right) = 0 \\
\Rightarrow {\left( {x + 2} \right)^2} = 0 \\
\therefore \left( {x + 2} \right) = 0 \\ \]
Therefore, we get the above value of x as -2.

Note: In quadratic equation, an alternative way of finding the factors is by directly solving the equation by using a direct formula which is given below.
\[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]
By substituting the values of a=1, b=4 and c=4 we get the factors of x.
\[x = \dfrac{{ - 4 \pm \sqrt {{{\left( 4 \right)}^2} - 4\left( 1 \right)\left( 4 \right)} }}{{2\left( 1 \right)}}\]
Therefore, the value we get of x is -2.