(9) The solution is x = 1 and x = 5
(10) The solution is x = 12 and x = -8
Explanation:
(9) The given expression is [tex](x+3)(x-3)=6 x-14[/tex]
We need to determine the solution of the equation.
The term [tex](x+3)(x-3)[/tex] can be expanded using the formula, [tex](a+b)(a-b)=a^{2}-b^{2}[/tex]
Thus, we have,
[tex]x^{2}-9=6 x-14[/tex]
[tex]x^{2}+5=6 x[/tex]
[tex]x^{2}-6 x+5=0[/tex]
Solving the quadratic equation, we get,
[tex](x-5)(x-1)=0[/tex]
The solution of the quadratic equation are x = 5 and x = 1
(10) The given expression is [tex]3 x^{2}-12 x-288=0[/tex]
We need to determine the solution of the equation.
Let us solve the equation using the quadratic formula,
[tex]x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}[/tex]
where [tex]a=3, b=-12, c=-288[/tex]
Thus, we have,
[tex]x=\frac{-(-12) \pm \sqrt{(-12)^{2}-4 \cdot 3(-288)}}{2 \cdot 3}[/tex]
Simplifying the terms, we have,
[tex]x=\frac{12 \pm \sqrt{144+3456}}{6}[/tex]
[tex]x=\frac{12 \pm \sqrt{3600}}{6}[/tex]
[tex]x=\frac{12 \pm60}{6}[/tex]
Thus, the roots are
[tex]x=\frac{12 +60}{6}[/tex] and [tex]x=\frac{12 -60}{6}[/tex]
[tex]x=\frac{72}{6}[/tex] and [tex]x=\frac{-48}{6}[/tex]
[tex]x=12[/tex] and [tex]x=-8[/tex]
Thus, the solution is x = 12 and x = -8
