Question

How do you solve $\left| {{x^2} + 6x} \right| = 3x + 18$ ?

Answer 1

Hint: We will consider the left-hand side of the given function in the form of negative and positive function separately as the modulus is given in the question. Then, factorize the polynomial using splitting the middle term method and find the required roots.

Complete step-by-step solution:
Given: Let $\left| {{x^2} + 6x} \right| = 3x + 18$ be the equation $(A)$
We can write the given expression as below
$\left| {{x^2} + 6x} \right| = - ({x^2} + 6x) ………... (1)$
And we can write $\left| {{x^2} + 6x} \right| = ({x^2} + 6x) ……..…. (2)$
Now, put all the values which we found in the equations $(1)$ and $(2)$ in the equation $(A)$ one by one.
So, we will be taking both the equations one by one to solve the roots.
$ - ({x^2} + 6x) = 3x + 18.........(3)$
Similarly,
$({x^2} + 6x) = 3x + 18.........(4)$
Hence, we got two equations $\left( 3 \right)$ and $\left( 4 \right)$ .
We are finding zeros of the equation $\left( 3 \right)$ and $\left( 4 \right)$
Now, we need to solve further,
In the equation $\left( 3 \right)$ ,
$ - {x^2} - 6x = 3x + 18$
Shifting the variable term to the left-hand side.
$\Rightarrow$$ - {x^2} - 6x - 3x = 18$
$\Rightarrow$$ - {x^2} - 9x = 18$
Because $( - )( - ) = ( + )$ and sign will be written of the bigger number
$\Rightarrow$$ - {x^2} - 9x - 18 = 0$
$\Rightarrow$$ - ({x^2} + 9x + 18) = 0$
$\Rightarrow$${x^2} + 9x + 18 = 0$
Using common factor method, write factor for $18 \Rightarrow {\text{18 = 2}} \times {\text{3}} \times {\text{3}}$
We will use the factors $6$ and $3$ .
Putting in the equation, we get,
$\Rightarrow$ ${x^2} + 6x + 3x + 18 = 0$
$\Rightarrow$$x(x + 6) + 3(x + 6) = 0$
$\Rightarrow$$(x + 6)(x + 3) = 0$
$\Rightarrow$$x + 6 = 0{\text{ or x + 3 = 0}}$
$\Rightarrow$$x = - 6$ or $x = - 3$
Similarly, we can solve the equation $(4)$
$\Rightarrow$$({x^2} + 6x) = 3x + 18$
Again, shifting the variables to the left-hand side,
$\Rightarrow$${x^2} + 6x - 3x = 18$
$\Rightarrow$${x^2} + 3x = 18$
$\Rightarrow$${x^2} + 3x - 18 = 0$
Using common factor method write factor for $ - 18 = ( - 2) \times 3 \times 6$
We will use the factors $6$ and $ - 3$ .
Using the factor method, we get,
$\Rightarrow$${x^2} + 6x - 3x - 18 = 0$
$\Rightarrow$$x\left( {x + 6} \right) - x\left( {x + 6} \right) = 0$
$\Rightarrow$$(x - 3)(x + 6) = 0$
To find the zeros of the polynomials, we can equate the factor equal to zero.
$\Rightarrow$$x - 3 = 0{\text{ or }}x + 6 = 0$
$\Rightarrow$$x = 3$ or $x = - 6$
Hence $3, - 6$ are the zeros of the polynomial.

We get the negative and the positive roots from the solving hence the answer would be $x = - 3, - 6{\text{ and 3}}{\text{.}}$

Note: Polynomials are the algebraic expressions which consist of variables and coefficients.
A factor is the number that divides into another number exactly.
To find the common factor of the two numbers, you first need to list all the factors of each one and then compare them.