Question

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

• A. 2
• B. 1
• C. 0
• D. 4

Hint:

For equations with square roots, ensure the expression inside is non-negative and the right-hand side is valid. Square both sides to solve, but verify solutions in the original equation.

Answer 1

The Correct Option is C

To solve the equation $\sqrt{3x^2 + x + 5} = x - 3$, we need to square both sides to eliminate the square root, but before doing that, let's consider the domain:

• For $\sqrt{3x^2 + x + 5}$ to be real, $3x^2 + x + 5 \geq 0$, which is true for all real numbers as it is a parabola opening upwards with no real roots.
• For $x - 3$ to be non-negative (since it's the RHS of a square root), $x \geq 3$.

Now, squaring both sides of the equation:

$3x^2 + x + 5 = (x - 3)^2$

Expanding the right-hand side:

$(x - 3)^2 = x^2 - 6x + 9$

Equating both sides, we get:

$3x^2 + x + 5 = x^2 - 6x + 9$

Simplifying this gives:

$2x^2 + 7x - 4 = 0$

To find the roots of the quadratic equation, we use the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = 2$, $b = 7$, and $c = -4$.

Calculating the discriminant:

$b^2 - 4ac = 7^2 - 4 \cdot 2 \cdot (-4) = 49 + 32 = 81$

Since the discriminant is positive, there are two real roots:

$x = \frac{-7 \pm 9}{4}$

Thus, the roots are:

$x = \frac{-7 + 9}{4} = \frac{2}{4} = \frac{1}{2}$

$x = \frac{-7 - 9}{4} = \frac{-16}{4} = -4$

Both roots need to satisfy $x \geq 3$, but:

• $x = \frac{1}{2}$ does not satisfy the condition $x \geq 3$.
• $x = -4$ does not satisfy the condition $x \geq 3$.

Therefore, there are no solutions that satisfy the original equation with the domain $x \geq 3$.

The number of solutions is 0.