If the set of all solutions of |x^2+x-9|=|x|+|x^2-9| is [ , ] [ ,), then ( ^2+ ^2+ ^2 ) is…

Question

If the set of all solutions of $|{x}^{2}+x-9|=|x|+|{x}^{2}-9|$ is $[\alpha ,\beta ]\cup [\gamma ,\infty)$, then ( ${\alpha }^{2}+{\beta }^{2}+{\gamma }^{2}$ ) is equal to:

Accepted Answer

We need to solve

$|{x}^{2}+x-9|=|x|+|{x}^{2}-9|.$

We will do it by considering intervals where the expressions inside modulus change sign.

The critical points are obtained from:

$x=0$ from $|x|$

$x=\pm 3$ from $|{x}^{2}-9|$

roots of ${x}^{2}+x-9=0$, but for NCERT-style interval splitting here, it is enough to use the main sign-change points $-3,0,3$ and then check sign of ${x}^{2}+x-9$ in each interval.

Now solve interval-wise.

1. For $x\le -3$
Here,

$|x|=-x$

$|{x}^{2}-9|={x}^{2}-9$

Also ${x}^{2}+x-9>0$, so $|{x}^{2}+x-9|={x}^{2}+x-9$

Thus,

${x}^{2}+x-9=-x+({x}^{2}-9)$

${x}^{2}+x-9={x}^{2}-x-9$

$x=-x$

$2x=0\Rightarrow x=0$

But $0-3$, so no solution here.

2. For $-3\le x\le 0$
Here,

$|x|=-x$

$|{x}^{2}-9|=9-{x}^{2}$

Now we check ${x}^{2}+x-9$ on this interval.

At $x=-3$,

${x}^{2}+x-9=9-3-9=-3At $x=0$,

${x}\^{2}+x-9=-9So throughout this interval,

$|{x}^{2}+x-9|=-({x}^{2}+x-9)=-{x}^{2}-x+9.$

Hence equation becomes

$-{x}^{2}-x+9=-x+(9-{x}^{2}).$

Simplify RHS:

$-{x}^{2}-x+9=-x+9-{x}^{2}.$

Both sides are identical, so this is true for every $x\in [-3,0]$.

So all $x\in [-3,0]$ are solutions.

3. For $0\le x\le 3$
Here,

$|x|=x$

$|{x}^{2}-9|=9-{x}^{2}$

Also on this interval,

${x}^{2}+x-9$

At $x=0$, it is $-90$.

So the sign changes inside this interval.

Let us find the root:

${x}^{2}+x-9=0$

$x=\frac{-1\pm \sqrt{37}}{2}.$

The positive root is

$\frac{-1+\sqrt{37}}{2}.$

So split further.

(i) For $0\le x\le \frac{-1+\sqrt{37}}{2}$
Here ${x}^{2}+x-9\le 0$, so

$|{x}^{2}+x-9|=-{x}^{2}-x+9.$

Equation:

$-{x}^{2}-x+9=x+(9-{x}^{2})$

$-{x}^{2}-x+9=-{x}^{2}+x+9$

$-x=x$

$2x=0\Rightarrow x=0.$

So only $x=0$, already included in previous interval.

(ii) For $\frac{-1+\sqrt{37}}{2}\le x\le 3$
Here ${x}^{2}+x-9\ge 0$, so

$|{x}^{2}+x-9|={x}^{2}+x-9.$

Equation:

${x}^{2}+x-9=x+(9-{x}^{2})$

${x}^{2}-9=9-{x}^{2}$

$2{x}^{2}=18$

${x}^{2}=9$

$x=\pm 3.$

In this interval, only $x=3$ is valid.

So from $[0,3]$, only $x=0,3$ are solutions.

4. For $x\ge 3$
Here,

$|x|=x$

$|{x}^{2}-9|={x}^{2}-9$

${x}^{2}+x-9>0$, so $|{x}^{2}+x-9|={x}^{2}+x-9$

Thus,

${x}^{2}+x-9=x+({x}^{2}-9)$

${x}^{2}+x-9={x}^{2}+x-9$

This is true for every $x\ge 3$.

So all $x\in [3,\infty)$ are solutions.

Complete solution set
Combining all intervals:

$[-3,0]\cup [3,\infty).$

Thus,

$\alpha =-3,\, \beta =0,\, \gamma =3.$

Now,

${\alpha }^{2}+{\beta }^{2}+{\gamma }^{2}=(-3{)}^{2}+{0}^{2}+{3}^{2}=9+0+9=18.$

So the correct answer is

$18$

Hence, Option B is correct.

Solutions · 1