Question

(1) \\(\sqrt{3x^2 + 7x + 10} = 4\\)

Explanation:

Step1: Eliminate the square root

Square both sides of the equation $\sqrt{3x^{2}+7x + 10}=4$ to get rid of the square root. According to the property that if $y = \sqrt{a}$, then $y^{2}=a$ (where $a\geq0$), we have $(\sqrt{3x^{2}+7x + 10})^{2}=4^{2}$, which simplifies to $3x^{2}+7x + 10 = 16$.

Step2: Transform into quadratic equation

Subtract 16 from both sides of the equation $3x^{2}+7x + 10 = 16$ to get a quadratic equation in standard form $ax^{2}+bx + c = 0$. So, $3x^{2}+7x+10 - 16=0$, which simplifies to $3x^{2}+7x - 6 = 0$.

Step3: Solve the quadratic equation

We can solve the quadratic equation $3x^{2}+7x - 6 = 0$ using the quadratic formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$, where $a = 3$, $b = 7$, and $c=- 6$. First, calculate the discriminant $\Delta=b^{2}-4ac=(7)^{2}-4\times3\times(-6)=49 + 72 = 121$. Then, $x=\frac{-7\pm\sqrt{121}}{2\times3}=\frac{-7\pm11}{6}$.

• For the plus sign: $x_{1}=\frac{-7 + 11}{6}=\frac{4}{6}=\frac{2}{3}$.
• For the minus sign: $x_{2}=\frac{-7-11}{6}=\frac{-18}{6}=- 3$.

Step4: Check the solutions

We need to check if these solutions satisfy the original square - root equation because squaring both sides of an equation can introduce extraneous solutions.

• For $x=\frac{2}{3}$:

Left - hand side of the original equation: $\sqrt{3\times(\frac{2}{3})^{2}+7\times\frac{2}{3}+10}=\sqrt{3\times\frac{4}{9}+\frac{14}{3}+10}=\sqrt{\frac{4}{3}+\frac{14}{3}+10}=\sqrt{\frac{18}{3}+10}=\sqrt{6 + 10}=\sqrt{16}=4$, which is equal to the right - hand side.

• For $x=-3$:

Left - hand side of the original equation: $\sqrt{3\times(-3)^{2}+7\times(-3)+10}=\sqrt{3\times9-21 + 10}=\sqrt{27-21 + 10}=\sqrt{16}=4$, which is equal to the right - hand side.

Answer:

The solutions of the equation $\sqrt{3x^{2}+7x + 10}=4$ are $x=\frac{2}{3}$ or $x=-3$.