Question

ii. solve and check for extran
④ $sqrt{(x + 2)^2-12}=2$.
$(x + 2)^2-12 = 4$
$sqrt{(x + 2)^2}=sqrt{16}$
$x + 2=4$ $x + 2=-4$
$\frac{-2 - 2}{x = 2}$ $\frac{-2 - 2}{x=-6}$
real solution real solution
iii. find a solution.
⑥ $3x + 2=sqrt{10 - x}$
① $5.4$ ② $6$ ③ $0.368$
iv. graph.
⑧ $f(x)=sqrt3{x + 3}-2$
① a graph with x - axis and y - axis, some points marked like - 2, - 3, 3

Explanation:

Solve $\sqrt{(x + 2)^2-12}=2$

Step1: Square both sides

Square both sides of the equation $\sqrt{(x + 2)^2-12}=2$ to get rid of the square - root. According to the property $(\sqrt{a})^2=a$ ($a\geq0$), we have $(x + 2)^2-12 = 4$.

Step2: Isolate the squared term

Add 12 to both sides of the equation $(x + 2)^2-12 = 4$. Then $(x + 2)^2=4 + 12=16$.

Step3: Take the square - root of both sides

Take the square - root of both sides of the equation $(x + 2)^2=16$. Remember that if $y^2=a$ ($a\geq0$), then $y=\pm\sqrt{a}$. So $x + 2=\pm4$.

Step4: Solve for $x$

Case 1: When $x + 2 = 4$, subtract 2 from both sides, we get $x=4 - 2=2$.
Case 2: When $x + 2=-4$, subtract 2 from both sides, we get $x=-4 - 2=-6$.

Step1: Square both sides

Square both sides of the equation $(3x + 2)^2=10 - x$. Expand the left - hand side using the formula $(a + b)^2=a^2+2ab + b^2$. So $(3x)^2+2\times3x\times2+2^2=10 - x$, which simplifies to $9x^2+12x + 4=10 - x$.

Step2: Rearrange to form a quadratic equation

Move all terms to one side to get a quadratic equation: $9x^2+12x+x+4 - 10=0$, or $9x^2+13x - 6=0$.

Step3: Use the quadratic formula

For a quadratic equation $ax^2+bx + c=0$ ($a
eq0$), the solutions are given by $x=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a}$. Here, $a = 9$, $b = 13$, and $c=-6$. First, calculate the discriminant $\Delta=b^2 - 4ac=(13)^2-4\times9\times(-6)=169 + 216=385$. Then $x=\frac{-13\pm\sqrt{385}}{18}$.
$x=\frac{-13+\sqrt{385}}{18}\approx\frac{-13 + 19.62}{18}=\frac{6.62}{18}\approx0.368$ and $x=\frac{-13-\sqrt{385}}{18}\approx\frac{-13 - 19.62}{18}=\frac{-32.62}{18}\approx - 1.81$.
We need to check for extraneous solutions by substituting back into the original equation.
When $x = 0.368$, $3\times0.368+2=1.104 + 2=3.104$ and $\sqrt{10 - 0.368}=\sqrt{9.632}\approx3.104$.
When $x=-1.81$, $3\times(-1.81)+2=-5.43+2=-3.43$, and $\sqrt{10-(-1.81)}=\sqrt{11.81}\approx3.44$. Since the left - hand side is negative and the right - hand side (square root) is non - negative for $x=-1.81$, $x=-1.81$ is an extraneous solution.

Answer:

$x = 2$ or $x=-6$

Solve $3x + 2=\sqrt{10 - x}$