MCQ
The equation $\sqrt {3 {x^2} + x + 5} = x - 3$ , where $x$ is real, has
  • no solution
  • B
    exactly one solution
  • C
    exactly two solution
  • D
    exactly four solution

Answer

Correct option: A.
no solution
a
Consider $\sqrt{3 x^{2}+x+5}=x-3$

Squaring both the sides, we get

$3 x^{2}+x+5=(x-3)^{2}$

$\Rightarrow 3 x^{2}+x+5=x^{2}+9-6 x$

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

$\Rightarrow 2 x^{2}+8 x-x-4=0$

$\Rightarrow 2 x(x+4)-1(x+4)=0$

$\Rightarrow x=\frac{1}{2}$ or $x=-4$

For $x=\frac{1}{2}$ and $x=-4$

$L.H.S.$ $\neq$ $R.H.S.$ of equation, $\sqrt{3 x^{2}+x+5}$ $=x-3$

Also, for every $x \in R, \mathrm{LHS} \neq \mathrm{RHS}$ of the given equation. 

$\therefore $ Given equation has no solution.

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Similar questions

The equation of the normal to the ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$ at the point $(a\cos \theta ,\;b\sin \theta )$ is
To fill $12$ vacancies there are $25$ candidates of which five are from scheduled caste. If $3$ of the vacancies are reserved for scheduled caste candidates while the rest are open to all, then the number of ways in which the selection can be made
A group consists of 4 couples in which each of the 4 persons have one wife each.In how many ways could they be arranged in a straight line such that the men and women occupy alternate positions:
Let $m, n \in N$ and $\operatorname{gcd}(2, n)=1$. If $30\left(\begin{array}{l}30 \\ 0\end{array}\right)+29\left(\begin{array}{l}30 \\ 1\end{array}\right)+\ldots+2\left(\begin{array}{l}30 \\ 28\end{array}\right)+1\left(\begin{array}{l}30 \\ 29\end{array}\right)= n .2^{ m }$ then $n + m$ is equal to (Here $\left.\left(\begin{array}{l} n \\ k \end{array}\right)={ }^{ n } C _{ k }\right)$
If the set $\left\{\operatorname{Re}\left(\frac{z-\bar{z}+z \bar{z}}{2-3 z+5 \bar{z}}\right): z \in C , \operatorname{Re}(z)=3\right\}$ is equal to the interval $(\alpha, \beta]$, then $24(\beta-\alpha)$ is equal to
$\mathop {\lim }\limits_{x \to 0} \,\,\cos \frac{1}{x}$
How many different committees of 5 can be formed from 6 men and 4 women on which exact 3 men and 2 women serve?
If $\omega $ is a cube root of unity, then a root of the equation $\left| {\begin{array}{*{20}{c}}{x + 1}&\omega &{{\omega ^2}}\\\omega &{x + {\omega ^2}}&1\\{{\omega ^2}}&1&{x + \omega }\end{array}} \right| = 0$ is
If the sides $A B, B C$ and $C A$ of a triangle $A B C$ have $3,5$ and $6$ interior points respectively, then the total number of triangles that can be constructed using these points as vertices, is equal to ....... .
If, $(\text{a}+1)\text{x}^2+2(\text{a}+1)\text{x}+(\text{a}-2)=0$ then, for what parameter of ‘a’ the given equation have real and distinct roots?