Question

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Answer 1

Question in English:-

What is the value of x in the equation $\small\sqrt{x^2+2x-3}+\sqrt{x^2-x}=\sqrt{5(x-1)}\ ?$

Solution:-

Given,

$\small\longrightarrow\sqrt{x^2+2x-3}+\sqrt{x^2-x}=\sqrt{5(x-1)}$

First let us find the domain of the equation.

Here the restrictions are,

• $\smallx^2+2x-3\geq0\quad\dots(i)$

• $\smallx^2-x\geq0\quad\dots(ii)$

• $\small5(x-1)\geq0\quad\dots(iii)$

[∵ √x is defined only for x ≥ 0.]

Considering (i),

$\small\longrightarrow x^2+2x-3\geq0$

$\small\longrightarrow(x+3)(x-1)\geq0$

$\small\Longrightarrow x\in(-\infty,\ -3]\cup[1,\ \infty)\quad\dots(1)$

Considering (ii),

$\small\longrightarrow x^2-x\geq0$

$\small\longrightarrow x(x-1)\geq0$

$\small\Longrightarrow x\in(-\infty,\ 0]\cup[1,\ \infty)\quad\dots(2)$

Considering (iii),

$\small\longrightarrow 5(x-1)\geq0$

$\small\longrightarrow x-1\geq0$

$\small\Longrightarrow x\in[1,\ \infty)\quad\dots(3)$

Taking (1) ∧ (2) ∧ (3) we get,

$\small\longrightarrow x\in[1,\ \infty)$

This is the domain. D = [1, ∞).

Now,

$\small\longrightarrow\sqrt{x^2+2x-3}+\sqrt{x^2-x}=\sqrt{5(x-1)}$

$\small\longrightarrow\sqrt{(x+3)(x-1)}+\sqrt{x(x-1)}-\sqrt{5(x-1)}=0$

$\small\longrightarrow\sqrt{x-1}\left(\sqrt{x+3}+\sqrt x-\sqrt5\right)=0$

This implies,

$\small\longrightarrow\sqrt{x-1}=0$

$\small\longrightarrow x-1=0$

$\small\longrightarrow x=1\quad\dots(4)$

or,

$\small\longrightarrow\sqrt{x+3}+\sqrt x-\sqrt5=0$

$\small\longrightarrow\sqrt{x+3}=\sqrt5-\sqrt x$

$\small\longrightarrow x+3=(\sqrt5-\sqrt x)^2$

$\small\longrightarrow x+3=5+x-2\sqrt{5x}$

$\small\longrightarrow2\sqrt{5x}=2$

$\small\longrightarrow\sqrt{5x}=1$

$\small\longrightarrow5x=1$

$\small\longrightarrow x=\dfrac{1}{5}\quad\dots(5)$

Taking (4) ∨ (5),

$\small\text{ \longrightarrow x\in\left\{\dfrac{1}{5},\ 1\right\} }$

Now this set has to be intersected with the domain D.

$\small\text{ \longrightarrow x\in\left\{\dfrac{1}{5},\ 1\right\}\cap D }$

$\small\text{ \longrightarrow x\in\left\{\dfrac{1}{5},\ 1\right\}\cap[1,\ \infty) }$

$\small\longrightarrow x\in\{1\}$

$\small\text{ \Longrightarrow\underline{\underline{x=1}} }$

Hence 1 is the value of x.