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2 years ago

Select the two values of x that are roots of this equation 2x^2+1=5x

Mathematics

1 answer:

Zanzabum2 years ago

8 0

Answer:

2.28 and 0.22

Step-by-step explanation:

2x²+1=5x

2x²-5x+1=0

this cannot be properly factorized

so let's use the almighty formula,

$- b + - \sqrt{( {b}^{2} } - 4ac) \: \times \frac{1}{2a}$

where a=2, b=-5, c= 1

=-(-5)±√{(-5)²-4x2x1} x 1/(2 x 2)

=5±√25-8 x1/4

=5±√17 x ¼

=(5±4.12)/4

=2.28. or. 0.22

roots=2.28 and 0.22

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3 years ago

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$\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ A(\stackrel{x_1}{7}~,~\stackrel{y_1}{4})\qquad M(\stackrel{x_2}{\frac{19}{2}}~,~\stackrel{y_2}{\frac{7}{2}})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ AM=\sqrt{\left( \frac{19}{2}-7 \right)^2+\left( \frac{7}{2}-4 \right)^2} \\\\\\ AM=\sqrt{\left( \frac{5}{2}\right)^2+\left( -\frac{1}{2} \right)^2}\implies \boxed{AM\approx 2.549509756796392} \\\\[-0.35em] ~\dotfill$

$\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ B(\stackrel{x_1}{10}~,~\stackrel{y_1}{6})\qquad M(\stackrel{x_2}{\frac{19}{2}}~,~\stackrel{y_2}{\frac{7}{2}}) \\\\\\ BM=\sqrt{\left( \frac{19}{2}-10 \right)^2+\left( \frac{7}{2}-6 \right)^2} \\\\\\ BM=\sqrt{\left( -\frac{1}{2}\right)^2+\left( -\frac{5}{2} \right)^2}\implies \boxed{BM\approx 2.549509756796392}$

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nevsk [136]

Answer:

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Step-by-step explanation:

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2 years ago

Select the two values of x that are roots of this equation 2x^2+1=5x​

Select the two values of x that are roots of this equation 2x^2+1=5x