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Hint: Solve for cases when x>0 and $x\le 0$ individually and remove extraneous roots(means which does not satisfy two cases). Use property that |x| = x when x>0 and |x| = -x when $x\le 0$. Find the sum of all the real roots found. Alternatively, you can plot the given function and find the points at which the graph of the given function intersects the x-axis.

“Complete step-by-step answer:”

We will solve the above question for two cases.

Case I: x>0

We have |x| = x

$\Rightarrow {{x}^{2}}+5x+6=0$

Using the quadratic formula which states that for quadratic equation $a{{x}^{2}}+bx+c=0$ the roots are $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$

Here a = 1, b = 5 and c = 6

Using the quadratic formula, we get

$\begin{align}

& x=\dfrac{-5\pm \sqrt{{{5}^{2}}-4\left( 1 \right)\left( 6 \right)}}{2\left( 1 \right)} \\

& \Rightarrow x=\dfrac{-5\pm \sqrt{25-24}}{2} \\

& \Rightarrow x=\dfrac{-5\pm \sqrt{1}}{2}=\dfrac{-5\pm 1}{2} \\

& \Rightarrow x=\dfrac{-5+1}{2}\text{ or }x=\dfrac{-5-1}{2} \\

& \Rightarrow x=-2\text{ or }x=-3 \\

\end{align}$

Since x>0 both x = -2 and x = -3 are extraneous roots.

Case II: $x\le 0$

We have |x| = -x

$\Rightarrow {{x}^{2}}-5x+6=0$

Here a = 1, b = -5 and c = 6

Using the quadratic formula, we get

$\begin{align}

& x=\dfrac{-\left( -5 \right)\pm \sqrt{{{\left( -5 \right)}^{2}}-4\left( 1 \right)\left( 6 \right)}}{2\left( 1 \right)} \\

& \Rightarrow x=\dfrac{5\pm \sqrt{25-24}}{2} \\

& \Rightarrow x=\dfrac{5\pm \sqrt{1}}{2}=\dfrac{5\pm 1}{2} \\

& \Rightarrow x=\dfrac{5+1}{2}\text{ or }x=\dfrac{5-1}{2} \\

& \Rightarrow x=3\text{ or }x=2 \\

\end{align}$

Since $x\le 0$, both x = 3 and x =2 are extraneous roots.

Hence the given equation has no real roots

Hence option D is correct.

Note: Alternative solution:

If we know the graph of f(x) then the graph of f(|x|) is the same as the graph f(x) for x>0 and for $x\le 0$ the graph is the mirror image taken in the y-axis.

Let $f(x)={{x}^{2}}+5x+6$

Then $f\left( \left| x \right| \right)={{x}^{2}}+5\left| x \right|+6$

First, we draw the graph of f(x)

Then using the above-mentioned fact, we draw the graph of f(|x|)

As is evident from the graph, the function f(|x|) does not have any root because it does not intersect the x-axis.

Hence the given equation ${{x}^{2}}+5\left| x \right|+6=0$ has no real roots.

“Complete step-by-step answer:”

We will solve the above question for two cases.

Case I: x>0

We have |x| = x

$\Rightarrow {{x}^{2}}+5x+6=0$

Using the quadratic formula which states that for quadratic equation $a{{x}^{2}}+bx+c=0$ the roots are $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$

Here a = 1, b = 5 and c = 6

Using the quadratic formula, we get

$\begin{align}

& x=\dfrac{-5\pm \sqrt{{{5}^{2}}-4\left( 1 \right)\left( 6 \right)}}{2\left( 1 \right)} \\

& \Rightarrow x=\dfrac{-5\pm \sqrt{25-24}}{2} \\

& \Rightarrow x=\dfrac{-5\pm \sqrt{1}}{2}=\dfrac{-5\pm 1}{2} \\

& \Rightarrow x=\dfrac{-5+1}{2}\text{ or }x=\dfrac{-5-1}{2} \\

& \Rightarrow x=-2\text{ or }x=-3 \\

\end{align}$

Since x>0 both x = -2 and x = -3 are extraneous roots.

Case II: $x\le 0$

We have |x| = -x

$\Rightarrow {{x}^{2}}-5x+6=0$

Here a = 1, b = -5 and c = 6

Using the quadratic formula, we get

$\begin{align}

& x=\dfrac{-\left( -5 \right)\pm \sqrt{{{\left( -5 \right)}^{2}}-4\left( 1 \right)\left( 6 \right)}}{2\left( 1 \right)} \\

& \Rightarrow x=\dfrac{5\pm \sqrt{25-24}}{2} \\

& \Rightarrow x=\dfrac{5\pm \sqrt{1}}{2}=\dfrac{5\pm 1}{2} \\

& \Rightarrow x=\dfrac{5+1}{2}\text{ or }x=\dfrac{5-1}{2} \\

& \Rightarrow x=3\text{ or }x=2 \\

\end{align}$

Since $x\le 0$, both x = 3 and x =2 are extraneous roots.

Hence the given equation has no real roots

Hence option D is correct.

Note: Alternative solution:

If we know the graph of f(x) then the graph of f(|x|) is the same as the graph f(x) for x>0 and for $x\le 0$ the graph is the mirror image taken in the y-axis.

Let $f(x)={{x}^{2}}+5x+6$

Then $f\left( \left| x \right| \right)={{x}^{2}}+5\left| x \right|+6$

First, we draw the graph of f(x)

Then using the above-mentioned fact, we draw the graph of f(|x|)

As is evident from the graph, the function f(|x|) does not have any root because it does not intersect the x-axis.

Hence the given equation ${{x}^{2}}+5\left| x \right|+6=0$ has no real roots.

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