The number of distinct real solutions of the equation x|x+4|+3|x+2|+10=0 is

Question

The number of distinct real solutions of the equation $x|x+4|+3|x+2|+10=0$ is

Accepted Answer

To find the number of distinct real solutions for the given equation, we need to understand how the expressions inside the absolute values behave. The absolute value expressions are $|x+4|$ and $|x+2|$. The signs of $x+4$ and $x+2$ change at $x=-4$ and $x=-2$. These two critical points divide the real number line into three intervals:

$x\ge -2$

$-4\le x
$x

Let's analyze the equation $x|x+4|+3|x+2|+10=0$ for each case separately.

Case 1: When $x\ge -2$

In this interval, both $x+4$ and $x+2$ are positive (or zero). So, $|x+4|=x+4$ and $|x+2|=x+2$.

Substitute these into the equation:

$x(x+4)+3(x+2)+10=0$

${x}^{2}+4x+3x+6+10=0$

${x}^{2}+7x+16=0$

Now, we check the discriminant $D$ (where $D={b}^{2}-4ac$) of this quadratic equation:

$D={7}^{2}-4(1)(16)=49-64=-15$

Since $DCase 2: When $-4\le x

Here, $x+4$ is positive (or zero), but $x+2$ is negative. So, $|x+4|=x+4$ and $|x+2|=-(x+2)$.

The equation becomes:

$x(x+4)-3(x+2)+10=0$

${x}^{2}+4x-3x-6+10=0$

${x}^{2}+x+4=0$

Let's check the discriminant for this equation:

$D={1}^{2}-4(1)(4)=1-16=-15$

Again, $DCase 3: When $x

In this case, both $x+4$ and $x+2$ are negative. So, $|x+4|=-(x+4)$ and $|x+2|=-(x+2)$.

The equation changes to:

$-x(x+4)-3(x+2)+10=0$

$-{x}^{2}-4x-3x-6+10=0$

$-{x}^{2}-7x+4=0$

Multiplying by $-1$ makes it easier to solve:

${x}^{2}+7x-4=0$

Let's find the discriminant:

$D={7}^{2}-4(1)(-4)=49+16=65$

Since $D>0$, this quadratic equation gives two real roots. Using the quadratic formula $x=\frac{-b\pm \sqrt{D}}{2a}$, the roots are:

$x=\frac{-7\pm \sqrt{65}}{2}$

Now we must verify if these roots actually lie in our current interval, which is $xWe know that $\sqrt{65}$ is slightly larger than $8$ (because $\sqrt{64}=8$). Let's roughly estimate it as $8.06$.

For the first root (taking the positive sign):

${x}\_{1}=\frac{-7+\sqrt{65}}{2}\approx \frac{-7+8.06}{2}\approx 0.53$

This value is greater than $-4$, so it does not lie in the given interval $xFor the second root (taking the negative sign):

${x}_{2}=\frac{-7-\sqrt{65}}{2}\approx \frac{-7-8.06}{2}\approx -7.53$

This value is strictly less than $-4$, so it falls correctly within our interval.

This means there is exactly $1$ valid real solution for the entire equation.

Option D is correct.

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