Question

How do you solve \[{x^2} + 6x + 8 = 0\] using the quadratic formula?

Answer 1

Hint: polynomial of degree two is called a quadratic polynomial and its zeros can be found using many methods like factorization, completing the square, graphs, quadratic formula etc. The quadratic formula is used when we fail to find the factors of the equation. In the given question we have to solve the given quadratic equation using the quadratic formula. That is \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\] .

Complete step-by-step answer:
Given, \[{x^2} + 6x + 8 = 0\] .
On comparing the given equation with the standard quadratic equation \[a{x^2} + bx + c = 0\] . We get, \[a = 1\] , \[b = 6\] and \[c = 8\] .
Substituting in the formula of standard quadratic, \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\] .
 \[ \Rightarrow x = \dfrac{{ - 6 \pm \sqrt {{6^2} - 4(1)(8)} }}{{2(1)}}\]
 \[ \Rightarrow x = \dfrac{{ - 6 \pm \sqrt {36 - 32} }}{2}\]
 \[ \Rightarrow x = \dfrac{{ - 6 \pm \sqrt 4 }}{2}\]
We know that the square root of 4 is 2.
 \[ \Rightarrow x = \dfrac{{ - 6 \pm 2}}{2}\]
 \[ \Rightarrow x = \dfrac{{2\left( { - 3 \pm 1} \right)}}{2}\]
 \[ \Rightarrow x = - 3 \pm 1\]
Thus there are two roots. That is
 \[ \Rightarrow x = - 3 + 1\] and \[x = - 3 - 1\]
 \[ \Rightarrow x = - 2\] and \[x = - 4\] .
Hence, the factors are \[x + 2\] and \[x + 4\] .
So, the correct answer is “\[x + 2\] and \[x + 4\]”.

Note: We can also solve this using simple factorization. That is the middle term ‘6x’ can be expanded as ‘4x’ and ‘2x’ (Because the multiplication of 4 and 2 is 8. If we multiply 4 and 2 we get 30 and if we add 4 and 2 we get 6). That is \[ \Rightarrow {x^2} + 6x + 8 = 0\]
 \[ \Rightarrow {x^2} + 4x + 2x + 8 = 0\]
Taking ‘x’ common on the first two terms and 2 common on the last two terms we have,
 \[ \Rightarrow x(x + 4) + 2(x + 4) = 0\]
Taking \[(x + 4)\] common on the above equation we have,
 \[ \Rightarrow (x + 4)(x + 2) = 0\] . That is the two factors are \[(x + 4)(x + 2)\] .