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2 years ago

X=square root of 880-8x

1 answer:

5 0

$x=\sqrt{880-8x}\\
D:880-8x\geq0 \wedge x\geq0\\
D:8x\leq880 \wedge x\geq0\\
D:x\leq 110 \wedge x\geq0\\
D:x\in\langle0,110\rangle\\
x^2=880-8x\\
x^2+8x-880=0\\
x^2+8x+16-896=0\\
(x+4)^2=896\\
x+4=8\sqrt{14} \vee x+4=-8\sqrt{14}\\
x=-4+8\sqrt{14} \vee x=-4-8\sqrt{14}\\
\\ x=-4-8\sqrt{14}\not \in D \Rightarrow \boxed{x=-4+8\sqrt{14}}$

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Serjik [45]

Given equation:

$(3x-4i)(7x-8)(3x+4i)=0$

i.e., $3x-4i=0$ and $7x-8=0$ and $3x+4i=0$

i.e., $3x=4i$ and $7x=8$ and $3x=-4i$

i.e., $x=\frac{4i}{3}$ and $x=\frac{8}{7}$ and $x=-\frac{4i}{3}$

Therefore, the given equation has a real root i.e., $x=\frac{8}{7}$

and two complex conjugate roots i.e., $x=\frac{4i}{3}$ and $x=-\frac{4i}{3}$

Hence, option C is correct.

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2 years ago

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natta225 [31]

Answer:

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Step-by-step explanation:

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mamaluj [8]

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Please Find the answer below

Step-by-step explanation:

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3 years ago

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Firlakuza [10]

Steps to solve:

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~Substitute

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~Simplify

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~Simplify

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~Add

~Subtract

Best of Luck!

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