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4-2.Quadratic Equations and Inequations
hard
The equation $\sqrt {3 {x^2} + x + 5} = x - 3$ , where $x$ is real, has
A
no solution
B
exactly one solution
C
exactly two solution
D
exactly four solution
(JEE MAIN-2014)
Solution
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.
Standard 11
Mathematics
