and graph the solution set on a number line. express the solution set in interval notation
x2 - 2x - 15 < 0
can anyone please help and explain this to me? i keep getting the wrong answer and its quite frustrating. please help! i would appreciate it soooo much!
x2 - 2x - 15 < 0
can anyone please help and explain this to me? i keep getting the wrong answer and its quite frustrating. please help! i would appreciate it soooo much!
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what you do is pretend that we have an equation.
solve:
x^2 - 2x - 15 = 0
factor:
(x - 5) (x + 3)= 0
x = 5 or x = -3.
now we unpretend we have an equation.
we put -3 and 5 on a number line, and we test a number less than -3, between -3 and 5, and more than 5 to see whether or not the number is positive or negative. we are looking for when it's less than 0.
----+..............-...............+--…
........-3..................5.........…
-4 is less than -3. lets plug x=-4 into the inequality.
(-4)^2 - 2 (-4) - 15 = 16 + 8 - 15 = 9 which is positive, so this is not right
0 is in between -3 and 5. lets check x=0
0^2 - 2(0) - 15 = -15. this is less than 0
6 is more than 5. lets check it.
6^2 - 2(6) - 15 = 36 - 12 - 15 = 9 . this is more than 0.
the solution for x2 - 2x - 15 < 0 is -3 < x < 5.
solve:
x^2 - 2x - 15 = 0
factor:
(x - 5) (x + 3)= 0
x = 5 or x = -3.
now we unpretend we have an equation.
we put -3 and 5 on a number line, and we test a number less than -3, between -3 and 5, and more than 5 to see whether or not the number is positive or negative. we are looking for when it's less than 0.
----+..............-...............+--…
........-3..................5.........…
-4 is less than -3. lets plug x=-4 into the inequality.
(-4)^2 - 2 (-4) - 15 = 16 + 8 - 15 = 9 which is positive, so this is not right
0 is in between -3 and 5. lets check x=0
0^2 - 2(0) - 15 = -15. this is less than 0
6 is more than 5. lets check it.
6^2 - 2(6) - 15 = 36 - 12 - 15 = 9 . this is more than 0.
the solution for x2 - 2x - 15 < 0 is -3 < x < 5.
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Sorry, this line editor does not accept graphs
x² – 2x – 15 < 0
First solve x² – 2x – 15 = 0, that is (x – 5)(x + 3) = 0, the solutions are x = 5 or x = -3
Since the graph of this polynomial is a parabola that opens up (I know the pattern, do you?) and the roots are -3 and 5, the domain [-3, 5] are values of x such that the polynomial is either on the x-axis or below the y-axis.
Therefore, the solution to your inequality is x in the open interval (-3, 5)
If you wish -3 < x < 5.
x² – 2x – 15 < 0
First solve x² – 2x – 15 = 0, that is (x – 5)(x + 3) = 0, the solutions are x = 5 or x = -3
Since the graph of this polynomial is a parabola that opens up (I know the pattern, do you?) and the roots are -3 and 5, the domain [-3, 5] are values of x such that the polynomial is either on the x-axis or below the y-axis.
Therefore, the solution to your inequality is x in the open interval (-3, 5)
If you wish -3 < x < 5.
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(x-5) (x+3) < 0
(x-5) < 0
add 5 to 0
x<5
(x+3) <0
subtract 3
x<-3
x< 5 or -3
(x-5) < 0
add 5 to 0
x<5
(x+3) <0
subtract 3
x<-3
x< 5 or -3