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FinnZ [79.3K]
2 years ago
12

Find all real zeros of tf(x)=x-14x² + 47x-18​

Mathematics
1 answer:
BabaBlast [244]2 years ago
4 0

Answer:

x=3/7 and x= 3

Step-by-step explanation:

First simplify:

-14 x^2 +48x -18

divide everything by -2

-2(7 x^2 -24x +9) (Multiply the leading coefficient 7 by the constant 9)

-2 (x^2 -24x +63)

-2 [(x-3)(x-21)] Now divide the 7 that you multiplied earlier from 3 and 21

-2 [(x-3/7)(x-21/7)]

-2 [(7x-3)(x-3)]

Hence, the zeroes are 3/7 and 3

Hope this helps!

If you think I helped, Please mark brainliest! Would really appreciate!

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5p - 14 = 8 p + 4 No solution, Infinitely many solutions, 6 ,-6​
DedPeter [7]

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-6

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3 years ago
<img src="https://tex.z-dn.net/?f=x%5E%7B2%7D%20-%2022x%20%2B121" id="TexFormula1" title="x^{2} - 22x +121" alt="x^{2} - 22x +12
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