<img src="https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B3%7Dx%5E%7B2%7D%20%2B5x%2B%5Cfrac%7B4%7D%7B3%7D%3D2x-%5Cfrac%7B1%7D%7B3%7D%20

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2 years ago

x%5E%7B2%7D" id="TexFormula1" title="\frac{1}{3}x^{2} +5x+\frac{4}{3}=2x-\frac{1}{3} x^{2}" alt="\frac{1}{3}x^{2} +5x+\frac{4}{3}=2x-\frac{1}{3} x^{2}" align="absmiddle" class="latex-formula">

Mathematics

1 answer:

Ber [7]2 years ago

5 0

X1= -4, x2= -1/2 is the correct answer

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Step-by-step explanation:

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Find the solution of 3 times the square root of the quantity of x plus 6 equals negative 12, and determine if it is an extraneou

Anna35 [415]

For this case we have the following equation:
$3 \sqrt{x+6}=-12$
Rewriting we have:
$\sqrt{x+6}= \frac{-12}{3}$
$\sqrt{x+6}=-4$
We raise both members of the equation to the square:
$(\sqrt{x+6})^2=(-4)^2$
Rewriting we have:
$x+6=16$
$x=16-6$
$x=10$
It's a extraneous solution because equality is not met by substituting x = 10 in the original equation:
$3 \sqrt{10+6}=-12$
$3 \sqrt{16}=-12$
$3(4)=-12$
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Answer:
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3 years ago

Will reward brainliest!

Lina20 [59]

Option D: Two irrational solutions

Explanation:

The equation is $17+3 x^{2}=6 x$

Subtracting 6x from both sides, we have,

$3x^{2} -6x+17=0$

Solving the equation using quadratic formula,

$x=\frac{6 \pm \sqrt{36-4(3)(17)}}{2(3)}$

Simplifying the expression, we get,

$\begin{aligned}x &=\frac{6 \pm \sqrt{36-204}}{6} \\&=\frac{6 \pm \sqrt{-168}}{6} \\&=\frac{6 \pm 2 i \sqrt{42}}{6}\end{aligned}$

Taking out the common terms and simplifying, we have,

$\begin{aligned}x &=\frac{2(3 \pm i \sqrt{42})}{6} \\&=\frac{(3 \pm i \sqrt{42})}{3}\end{aligned}$

Dividing by 3, we get,

$x=1+i \sqrt{\frac{14}{3}}, x=1-i \sqrt{\frac{14}{3}}$

Hence, the equation has two irrational solutions.

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Simplify the radical expression by rationalizing the denominator.9 square root of 25 divided by the square root of 50

Crank

(9√25) /√50 = 9*5/√50 now simplify the denominator. √50=√25*√2=5√2
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2 years ago

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