Answer:
[tex][0,9][/tex]
Step-by-step explanation:
We have been given an inequality [tex]x^2-6x+c<0[/tex]. We are asked to find the positive values of 'c' such that the given inequality has real solutions for x.
We will use discriminant formula to solve our given problem.
[tex]D=b^2-4ac[/tex]
For real solutions discriminant should be greater than or equal to 0 that is [tex]b^2-4ac\geq 0[/tex]
We can see that [tex]a=1[/tex] and [tex]b=-6[/tex] for our given problem, so we will get:
[tex](-6)^2-4(1)c\geq 0[/tex]
[tex]36-4c\geq 0[/tex]
[tex]36-36-4c\geq 0-36[/tex]
[tex]-4c\geq -36[/tex]
Dividing by a negative number will swap the inequality as:
[tex]\frac{-4c}{-4}\leq \frac{-36}{-4}[/tex]
[tex]c\leq 9[/tex]
Therefore, the value of 'c' would be less than or equal to 9.
Since we are asked to find the positive values of 'c', so the positive values of 'c' will begin at 0.
Therefore, our required interval would be [tex][0,9][/tex], which will give the real solutions for x.
