The equation sqrt{3 x^{2}+x+5}=x-3,where x is | vedclass

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(A) Consider the equation $\sqrt{3 x^{2}+x+5}=x-3$.For the square root to be defined and equal to a real number,the right-hand side must be non-negati

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(A) Consider the equation $\sqrt{3 x^{2}+x+5}=x-3$.For the square root to be defined and equal to a real number,the right-hand side must be non-negative,i.e.,$x-3 \geq 0$,which implies $x \geq 3$.Squaring both sides of the equation:$3 x^{2}+x+5=(x-3)^{2}$$3 x^{2}+x+5=x^{2}-6 x+9$$2 x^{2}+7 x-4=0$Factoring the quadratic equation:$2 x^{2}+8 x-x-4=0$$2 x(x+4)-1(x+4)=0$$(2 x-1)(x+4)=0$This gives $x = \frac{1}{2}$ or $x = -4$.Checking these values against the condition $x \geq 3$:Neither $x = \frac{1}{2}$ nor $x = -4$ satisfies the condition $x \geq 3$.Therefore,the given equation has no real solution.

The equation $\sqrt{3 x^{2}+x+5}=x-3$,where $x$ is real,has

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