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3 years ago

Solve for x in the equation 3 x squared minus 18 x + 5 = 47.

Mathematics

2 answers:

bekas [8.4K]3 years ago

7 0

Answer:

A. X = +/- sqr root 23

Step-by-step explanation:

I did the Edgenuity

ryzh [129]3 years ago

6 0

Answer:

x= 3 + $\sqrt{23}$ , 3 - $\sqrt{23}$

Step-by-step explanation:

You use the quadratic formula to get x= $\frac{18+6\sqrt{23} }{6} ,\frac{18-6\sqrt{23} }{6}$

Then you simplify and get the answers x= 3 + $\sqrt{23}$ , 3 - $\sqrt{23}$

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oksian1 [2.3K]

Answer:

<h3> 1) $x_1=\dfrac{7+\sqrt{13}}2\,,\quad x_2=\dfrac{7-\sqrt{13}}2$ </h3><h3> 2) $x_1=-\dfrac13\,,\quad x_2=-3$ </h3>

Step-by-step explanation:

$x^2 - 7x + 9 = 0\quad\implies\quad a=1\,,\ b = -7\,,\ c=9\\\\x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}=\dfrac{-(-7)\pm\sqrt{(-7)^2-4\cdot1\cdot9}}{2\cdot1}=\dfrac{7\pm\sqrt{49-36}}2\\\\x_1=\dfrac{7+\sqrt{13}}2\,,\quad x_2=\dfrac{7-\sqrt{13}}2$

<h3>2)</h3><h3> $3x^2 + 10x=-3\\\\3x^2+10x+3=0\quad\implies\quad a=3\,,\ b =10\,,\ c=3\\\\x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}=\dfrac{-10\pm\sqrt{10^2-4\cdot3\cdot3}}{2\cdot3}= \dfrac{-10\pm\sqrt{100-36}}6\\\\x_1=\dfrac{-10+\sqrt{64}}6=\dfrac{-10+8}6=-\dfrac13\,,\qquad x_2=\dfrac{-10-8}6=-3$ </h3>

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3 years ago

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Paladinen [302]

Answer:

x=2.75

Step-by-step explanation:

The answer will be correct.

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2 years ago