Question

solution set of 3|x-1| +x² -7>0.
no irrelevant answer!!​

Answer 1

Answer:

X>2

Step-by-step explanation:

3(x-1)+ x2-7

=3x-3+x2-7

=x2+3x-10

=x2+5x-2x-10

=x(x+5)-2(x+5)

=(x-2)(x+5)

X>2

Answer 2

We need to find the solution set of the inequality,

$\longrightarrow 3|x-1|+x^2-7>0$

Case 1:- Let,

• $x-1\geq0$

• $x\in[1,\ \infty)\quad\quad\dots(1.1)$

such that $|x-1|=x-1.$

Then,

$\longrightarrow 3(x-1)+x^2-7>0$

$\longrightarrow 3x-3+x^2-7>0$

$\longrightarrow x^2+3x-10>0$

$\longrightarrow x^2+5x-2x-10>0$

$\longrightarrow x(x+5)-2(x+5)>0$

$\longrightarrow(x-2)(x+5)>0$

$\Longrightarrow x\in(-\infty,\ -5)\cup(2,\ \infty)\quad\quad\dots(1.2)$

Taking $(1.1)\land(1.2),$

$\longrightarrow x\in[1,\ \infty)\cap\left[(-\infty,\ -5)\cup(2,\ \infty)\right]$

$\longrightarrow x\in(2,\ \infty)\quad\quad\dots(1)$

Case 2:- Let,

• $x-1\leq0$

• $x\in(-\infty,\ 1]\quad\quad\dots(2.1)$

such that $|x-1|=1-x.$

Then,

$\longrightarrow 3(1-x)+x^2-7>0$

$\longrightarrow 3-3x+x^2-7>0$

$\longrightarrow x^2-3x-4>0$

$\longrightarrow x^2-4x+x-4>0$

$\longrightarrow x(x-4)+x-4>0$

$\longrightarrow(x+1)(x-4)>0$

$\Longrightarrow x\in(-\infty,\ -1)\cup(4,\ \infty)\quad\quad\dots(2.2)$

Taking $(2.1)\land(2.2),$

$\longrightarrow x\in(-\infty,\ 1]\cap\left[(-\infty,\ -1)\cup(4,\ \infty)\right]$

$\longrightarrow x\in(-\infty,\ -1)\quad\qua\dots(2)$

Finally $(1)\lor(2)$ gives the solution set of our inequality.

$\longrightarrow\underline{\underline{x\in(-\infty,\ -1)\cup(2,\ \infty)}}$