Question

`\huge\bold\red{{HELP}}`​

Answer 1

1)

We note that the quadratic can be factored into
`(2x-1)(x-1)`.

The quadratic is greater than
`0` if both of its factors are positive, or they are both negative.

Case: both positive

We need to solve the system of inequalities:

`2x-1>0\\x-1>0`.

The first inequality gives
`x>(1)/(2)`.

The second inequality gives
`x>1`.

Taking the points where the inequalities coincide gives
`x>1`.

(Note: 1 is a root of the quadratic. Coincidence? If not, try and prove it!)

Case: both negative

We need to solve the system of inequalities:

`2x-1<0\\x-1<0`.

The first inequality gives
`x<(1)/(2)`.

The second inequality gives
`x<1`.

Taking the points where the inequalities coincide gives
`x<(1)/(2)`.

(Note:
`(1)/(2)` is the other root of the quadratic. Coincidence? If not, try and prove it!)

Taking the union of both cases gives the solution set:
`\boxed{x\in (-\infty, (1)/(2)) \cup (1, \infty)}`

2)

We bring over the
`15` to get
`x^2-2x-15<0`.

Note that the quadratic factors into
`(x-5)(x+3)`.

The quadratic is less than
`0` if 1 of its factors is negative, but not both.

Case: first factor is negative, second positive

We have that
`x-5<0` and
`x+3>0`.

We get that
`x<5` and
`x>-3`, which has the solution set
`-3<x<5`.

Case: second factor is negative, first positive

We have that
`x+3<0` and
`x-5>0`.

We get that
`x<-3` and
`x>5`, which has no solutions.

So, the solution set is
`\boxed{x\in (-3, 5)}`