Solve h(x)=x^2-2x-8>7

Mathematics

1 answer:

PIT_PIT [208]3 years ago

7 0

Answer:

x < -3 and x > 5

Step-by-step explanation:

x² - 2x - 8 > 7

~Rewrite in standard form

x² - 2x - 15 > 0

~Factor

(x + 3)(x - 5) = 0

~Solve for both factors

x + 3 = 0

x = -3

x - 5 = 0

x = 5

Best of Luck!

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Find the distance between the points of intersection of the graphs of the functions.

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Answer:

Find the value of x and y in coordinate form, that'll be the point of intersection.

Question 1

${ \rm{y = {x}^{2} - 3x + 4 }} \\ { \boxed{ \tt{but \: y = x + 1 \: }}} \\ \\ { \rm{(x + 1) = {x}^{2} - 3x + 4 }} \\ \\ { \rm{ {x}^{2} - 4x + 3 = 0 }} \\ \\ { \rm{(x - 3)(x - 1) = 0}} \\ \\ { \boxed{ \rm{x_{1} = 3 \: \: and \: \: x _{2} = 1}}} \\ \\ { \boxed{ \tt{remember \: y = x + 1}}} \\ \\ { \rm{y _{1} = 4 \: \: and \: \: y _{2} = 2 }}$

Therefore, points of intersection are two

Answer:  (3, 4) and (1, 2)

Question 2:

Following the steps as in question 1

${ \rm{y = {x}^{2} - 4}} \\ \\{ \rm{2x - 4 = {x}^{2} - 4}} \\ \\ { \rm{ {x}^{2} = 2x }} \\ \\ { \boxed{ \rm{x = 2}}} \\ { \tt{remember : \: y = 2x - 4 }} \\ { \boxed{ \rm{y = 0}}}$