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marissa [1.9K]
3 years ago
6

Solve for x.6x^2=4x+10

Mathematics
1 answer:
Agata [3.3K]3 years ago
6 0

Answer:

x=-1,\displaystyle \frac{5}{3}

Step-by-step explanation:

Hi there!

6x^2=4x+10

Move 6x^2 to the other side of the equation:

0=-6x^2+4x+10

Divide both sides by -2:

0=3x^2-2x-5

Factor by grouping:

0=3x^2+3x-5x-5\\0=3x(x+1)-5(x+1)\\0=(3x-5)(x+1)

Apply the zero product property:

3x-5 = 0\\3x=5\\\\x=\displaystyle \frac{5}{3}

x+1=0\\x=-1

Therefore, x = -1, 5/3.

I hope this helps!

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Check the picture below.

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\bf A(\stackrel{x_1}{1}~,~\stackrel{y_1}{4})\qquad B(\stackrel{x_2}{5}~,~\stackrel{y_2}{1}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{1}-\stackrel{y1}{4}}}{\underset{run} {\underset{x_2}{5}-\underset{x_1}{1}}}\implies \cfrac{-3}{4} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{slope of AB}}{-\cfrac{3}{4}}\qquad \qquad \qquad \stackrel{\textit{\underline{negative reciprocal} and slope of the diameter}}{\cfrac{4}{3}}

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\bf ~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ A(\stackrel{x_1}{1}~,~\stackrel{y_1}{4})\qquad B(\stackrel{x_2}{5}~,~\stackrel{y_2}{1}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left( \cfrac{5+1}{2}~~,~~\cfrac{1+4}{2} \right)\implies \left(3~~,~~\cfrac{5}{2} \right)

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\bf (\stackrel{x_1}{3}~,~\stackrel{y_1}{\frac{5}{2}}) \stackrel{slope}{m}\implies \cfrac{4}{3} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{\cfrac{5}{2}}=\stackrel{m}{\cfrac{4}{3}}(x-\stackrel{x_1}{3})\implies y-\cfrac{5}{2}=\cfrac{4}{3}x-4 \\\\\\ y=\cfrac{4}{3}x-4+\cfrac{5}{2}\implies y=\cfrac{4}{3}x-\cfrac{3}{2}

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3 years ago
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