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3 years ago

6

If f(x)=2x^2-x-6 and g(x)=x^2-4 find f(x)/g(x)

1 answer:

DENIUS [597]3 years ago

6 0

$\dfrac{f(x)}{g(x)}=\dfrac{2x^2-x-6}{x^2-4}\\
\dfrac{f(x)}{g(x)}=\dfrac{2x^2-4x+3x-6}{(x-2)(x+2)}\\
\dfrac{f(x)}{g(x)}=\dfrac{2x(x-2)+3(x-2)}{(x-2)(x+2)}\\
\dfrac{f(x)}{g(x)}=\dfrac{(2x+3)(x-2)}{(x-2)(x+2)}\\
\dfrac{f(x)}{g(x)}=\dfrac{2x+3}{x+2}$

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3 years ago

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Vedmedyk [2.9K]

Answer:

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and there are no real roots of the equation given above

Step-by-step explanation:

To solve:

5x² − 3x + 17 = 9

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⇒ 5x² − 3x + 17 - 9 = 0

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⇒ 5x² − 3x + 8 = 0

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the roots of the equation in the form ax² + bx + c = 0 is given as:

x = $\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

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c = 8

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here i = √(-1)

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