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3 years ago

Solve for x. Help, please. ༼ つ ◕_◕ ༽つ

Mathematics

1 answer:

ahrayia [7]3 years ago

5 0

Answer:

x = -10

Step-by-step explanation:

first you multiply both sides by -3

x + 7 = -3

then you subtract 7 from both sides

x = -10

hope this helps

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3 years ago

Solve for the roots in the equation below. In your final answer, include each of the necessary steps and calculations. Hint: Use

QveST [7]

$\bf 27\implies 3^3\\\\
i^3\implies i\cdot i\cdot i\implies i^2\cdot i\implies -1\cdot i\implies -i\\\\
-----------------------------\\\\
x^3-27i=0\implies x^3+3^3(-i)=0\implies x^3+(3^3i^3)=0
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x^3+(3i)^3=0\\\\
-----------------------------\\\\
\textit{difference of cubes}
\\ \quad \\
a^3+b^3 = (a+b)(a^2-ab+b^2)\qquad
(a+b)(a^2-ab+b^2)= a^3+b^3\\\\
-----------------------------\\\\$

$\bf (x+3i)[x^2-3ix+(3i)^2]=0\implies 
\begin{cases}
x+3i=0\implies \boxed{x=-3i}\\\\
x^2-3ix+(3i)^2=0
\end{cases}
\\\\\\
\textit{now, for the second one, we'd need the quadratic formula}
\\\\\\
x^2-3ix+(3i)^2=0\implies x^2-3ix+(3^2i^2)=0
$

$\bf x^2-3ix+(-9)=0\implies x^2-3ix-9=0
\\\\\\
\textit{quadratic formula}\\\\
x= \cfrac{ - {{ b}} \pm \sqrt { {{ b}}^2 -4{{ a}}{{ c}}}}{2{{ a}}}\implies x=\cfrac{3i\pm\sqrt{(-3i)^2-4(1)(-9)}}{2(1)}
\\\\\\
x=\cfrac{3i\pm\sqrt{(-3)^2i^2+36}}{2}\implies x=\cfrac{3i\pm\sqrt{-9+36}}{2}
\\\\\\
x=\cfrac{3i\pm\sqrt{27}}{2}\implies \boxed{x=\cfrac{3i\pm 3\sqrt{3}}{2}}$

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