Question

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 solutions
(d) exactly four solutions

Answer 1

The given equation in the above question is in the form of square root, which gives only non-negative values. From this, we can write the condition $x\ge 3$. Then we have to solve the given equation. For this, we need to take the square on both the sides of the given equation to obtain the equation $3{{x}^{2}}+x+5={{\left( x-3 \right)}^{2}}$ which can be simplified using the algebraic identity ${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$ to the standard quadratic equation form given by $a{{x}^{2}}+bx+c=0$. Then using the quadratic formula, given by $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$, we can solve obtain the solutions. Finally, using the condition $x\ge 3$ we can reject the incorrect solutions and obtain the required number of solutions.

Complete step by step solution:
The equation given in the above question is
$\Rightarrow \sqrt{3{{x}^{2}}+x+5}=x-3........\left( i \right)$
We know that the square root function gives only the non-negative values. So the term on the RHS of the above equation must also be non-negative, which means that
$\begin{aligned} & \Rightarrow x-3\ge 0 \\\ & \Rightarrow x\ge 3........\left( ii \right) \\\ \end{aligned}$
Taking squares of both the sides of the equation (i) we get

& \Rightarrow {{\left( \sqrt{3{{x}^{2}}+x+5} \right)}^{2}}={{\left( x-3 \right)}^{2}} \\\ & \Rightarrow 3{{x}^{2}}+x+5={{\left( x-3 \right)}^{2}} \\\ \end{aligned}$$ Now, using the algebraic identity $${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$$ we can expand the RHS of the above equation as $$\begin{aligned} & \Rightarrow 3{{x}^{2}}+x+5={{x}^{2}}-2\left( 3 \right)\left( x \right)+{{3}^{2}} \\\ & \Rightarrow 3{{x}^{2}}+x+5={{x}^{2}}-6x+9 \\\ \end{aligned}$$ Subtracting $${{x}^{2}}-6x+9$$ from both the sides, we get $$\begin{aligned} & \Rightarrow 3{{x}^{2}}+x+5-\left( {{x}^{2}}-6x+9 \right)={{x}^{2}}-6x+9-\left( {{x}^{2}}-6x+9 \right) \\\ & \Rightarrow 3{{x}^{2}}+x+5-{{x}^{2}}+6x-9=0 \\\ & \Rightarrow 2{{x}^{2}}+7x-4=0.......\left( i \right) \\\ \end{aligned}$$ By comparing the above equation with the standard form of a quadratic equation, $${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$$, we get $\begin{aligned} & \Rightarrow a=2.......\left( ii \right) \\\ & \Rightarrow b=7.......\left( iii \right) \\\ & \Rightarrow c=-4.......\left( iv \right) \\\ \end{aligned}$ Now, we know that the quadratic formula is given by $\Rightarrow x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ Substituting (ii), (iii), and (iv) in the above equation we get $$\begin{aligned} & \Rightarrow x=\dfrac{-7\pm \sqrt{{{\left( 7 \right)}^{2}}-4\left( 2 \right)\left( -4 \right)}}{2\left( 2 \right)} \\\ & \Rightarrow x=\dfrac{-7\pm \sqrt{49+32}}{4} \\\ & \Rightarrow x=\dfrac{-7\pm \sqrt{81}}{4} \\\ & \Rightarrow x=\dfrac{-7\pm 9}{4} \\\ \end{aligned}$$ On simplifying we get $\begin{aligned} & \Rightarrow x=\dfrac{2}{4},x=\dfrac{-16}{4} \\\ & \Rightarrow x=\dfrac{1}{2},x=-4 \\\ \end{aligned}$ But from (ii), $x\ge 3$ which none of the above two solutions satisfied. Thus, the given equation has no solution. **Hence, the correct answer is option (a).** **Note:** Do not forget to obtain the constraint $x\ge 3$ before proceeding to solve the given equation. Always remember to obtain such constraints whenever the square root function is there in the equations. We may argue that the square root is defined only for non-negative values, which means $3{{x}^{2}}+x+5\ge 0$. We can easily prove this by checking the discriminant of this quadratic polynomial, which will come out to be negative.