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

Mathematics

1 answer:

BabaBlast [244]3 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

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A number line goes from negative 5 to positive 5. Point D is at negative 4 and point E is at positive 5. A line is drawn from po

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Directed numbers are numbers that have either a positive or negative sign, which can be shown on a number line. Therefore, point F is Fifteen-halves of line segment DE.

A number line is a system that can show the positions of positive or negative numbers. It has its ends ranging from negative infinity to positive infinity. Thus any directed number can be located on the line.

Directed numbers are numbers with either a negative or positive sign, which shows their direction with respect to the number line.

In the given question, the distance between points D and E is 9 units. So that dividing 9 units in the ratio of 5 to 6, we have;

$\frac{5}{6}$ x 9 = $\frac{45}{6}$

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Therefore, the location of point F, which partitions the directed line segment from d to E into a 5:6 ratio is $\frac{15}{2}$ . Thus the answer is Fifteen-halves.

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