29 Jun, 2021 03:15 PM

how would i solve this?

I'm having a lot of trouble understanding math, some help would be appreciative. the question is to determine zeros from following equations: a) g(x) = 2x^2 - x - 6 b) h(x) = 2^x - 1 c) j(x) = sin (x - 45degrees), where 0degrees < or = x < or = 360degrees d) k(x) = 2 cosx, where -360degrees < or = x < or = 0degrees

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Answer
FI
29 Jun, 2021 03:15 PM

a) g(x) = 2x^2 - x - 6 

Ans:

The zeroes of g(x) is given when,

2x^2 - x - 6 = 0

2x^2 - 4x + 3x - 6 = 0

2x(x - 2) + 3(x - 2) = 0

(2x + 3)(x - 2) = 0

2x + 3 = 0

2x = -3

x = -3/2 or

x - 2 = 0

x = 2

The zeroes are when x = -3/2 and 2

b) h(x) = 2^x - 1 

Ans:

2x^2 - 1 = 0

2x^2 = 1

x^2 = 1/2

x = ±√(1/2)

The zeroes are when x = -√(1/2) and +√(1/2)

c) j(x) = sin (x - 45degrees), where 0degrees < or = x < or = 360degrees 

Ans:

When j(x) = 0

Sin (x - 45) = 0

Sin^-1 (0) = x - 45

0 = x - 45

45 = x

Recall sin a = sin (180 - a)

x - 45 = 180

x = 180 + 45

x = 225

Thus, the zeroes are when x = 45 degrees and 225 degrees

d) k(x) = 2 cosx, where -360degrees < or = x < or = 0degrees

Ans:

When k(x) = 0,

2cosx = 0

Cosx = 0/2 = 0

Cosx = 0

Cos^-1 0 = x

x = 90