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TSI Mathematics Videos 144 videos

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TSI Math: Finding a Range of Solutions for a Variable in a Triangle's Equation 45 Views


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Description:

What is the range of all possible values of x for ∆DEF, if DE = 2x – 7, EF = 3x + 9, and DF = 2x + 5?


Transcript

00:02

Okay next time for shoppers What is the range of

00:05

all possible values of x for angle df if d

00:10

is to explain a seven e f is three x

00:13

plus nine and gf is two x plus five Okay

00:20

interesting problem here Three lines have met forming a triangle

00:24

and psych test takers everywhere are scrambling to determine a

00:27

little fact about this unholy union Like all heroes intent

00:31

on saving the universe from evil we need a strategy

00:33

Our strategy relies on the triangle inequality serum It says

00:37

that some of the lengths of any two sides of

00:39

a triangle must be greater than the length of the

00:42

third side Right Otherwise the triangle wouldn't touch so with

00:45

angle df we know that d plus e s greater

00:48

than d f e f plus d f is greater

00:50

than d e n d f plus d e is

00:52

greater than e f right So all the little side

00:55

thing he's touch will use this knowledge along with the

00:57

more algebraic definitions of the seides find all the possible

01:00

ranges of x let's start with d plus c f

01:03

his great of ndf and we'll get to x minus

01:05

seven plus three X plus nine is greater than two

01:08

x plus five But when we solve this inequality we

01:11

end up with yes x is greater than one right

01:14

Actually did all the math get point that x is

01:16

greater Next we have df plus cf scarier than d

01:19

e that gives us two x plus five plus two

01:21

extra stein's greater than to explain it's Seven solve it

01:24

and we get axes while greater than what is that

01:26

negative seven Yeah well this is obvious because we already

01:29

know that x is greater than once Of course it

01:31

has to be great the negative seven as well We

01:33

can't have a triangle with negative sides at least not

01:36

in this reality So lastly let's solve df plus de

01:39

is greater than e f and we get two x

01:41

plus five plus two extra money Seven is greater than

01:44

three x plus nine This will give us access great

01:46

event while eleven This inequality trumps the other two and

01:50

is the only one that matters and the value of

01:52

x can be anything It wants to be as long

01:55

as it's larger than eleven So the answer is d 00:01:57.456 --> [endTime] and yes we are shmoop oh snap

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