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

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(A) Given the equation $\sqrt{3x^2 + x + 5} = x - 3$.For the square root to be defined and equal to a real value,we must have $x - 3 \geq 0$,which imp

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(A) Given the equation $\sqrt{3x^2 + x + 5} = x - 3$.For the square root to be defined and equal to a real value,we must have $x - 3 \geq 0$,which implies $x \geq 3$.Squaring both sides:$3x^2 + x + 5 = (x - 3)^2$$3x^2 + x + 5 = x^2 - 6x + 9$$2x^2 + 7x - 4 = 0$Solving the quadratic equation $2x^2 + 7x - 4 = 0$ using the quadratic formula or factorization:$2x^2 + 8x - x - 4 = 0$$2x(x + 4) - 1(x + 4) = 0$$(2x - 1)(x + 4) = 0$So,$x = \frac{1}{2}$ or $x = -4$.Checking these values against the condition $x \geq 3$:For $x = \frac{1}{2}$,$\frac{1}{2} For $x = -4$,$-4 Since neither value satisfies the condition $x \geq 3$,the equation has no real solution.

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

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