The number of solutions of the equation sqrt{ | vedclass

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(C) Given the equation $\sqrt{3x^2+x+5} = x-3$. Squaring both sides,we get $3x^2+x+5 = (x-3)^2$. Expanding the right side: $3x^2+x+5 = x^2-6x+9$. Rear

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(C) Given the equation $\sqrt{3x^2+x+5} = x-3$. Squaring both sides,we get $3x^2+x+5 = (x-3)^2$. Expanding the right side: $3x^2+x+5 = x^2-6x+9$. Rearranging terms: $2x^2+7x-4 = 0$. Factoring the quadratic: $2x^2+8x-x-4 = 0$,which gives $2x(x+4)-1(x+4) = 0$. So,$(2x-1)(x+4) = 0$,leading to $x = 1/2$ or $x = -4$. Now,we must check these values in the original equation $\sqrt{3x^2+x+5} = x-3$. For $x = 1/2$: $\sqrt{3(1/4)+1/2+5} = \sqrt{0.75+0.5+5} = \sqrt{6.25} = 2.5$,while $x-3 = 0.5-3 = -2.5$. Since $2.5 
eq -2.5$,$x = 1/2$ is not a solution. For $x = -4$: $\sqrt{3(16)-4+5} = \sqrt{48+1} = \sqrt{49} = 7$,while $x-3 = -4-3 = -7$. Since $7 
eq -7$,$x = -4$ is not a solution. Thus,there are no real solutions to the equation.

The number of solutions of the equation $\sqrt{3x^2+x+5} = x-3$ is

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