Question

Can someone please help with these 3 I am so confused on what to do thank you!!!!

Answer 1

26.
`(3)/(4) \leq x <7`

27.
`x\leq 5`

28.
`0\leq b<4`

Explanation:

26.
`√(4x-3) <5`:

Taking square on both the sides to get:

`(√(4x-3) )^2 < (5)^2`

`4x-3<25`

`4x<28`

`x<(28)/(4)`

`x<7`

For non-negative values for radical:

`4x-3\geq 0`

`x\geq (3)/(4)`

So solution for this:
`(3)/(4) \leq x <7`

27.
`2+√(4x-4)\leq  6`

Subtracting 2 from both the sides to get:

`2+√(4x-4)-2\leq  6-2`

`√(4x-4)\leq 4`

Taking square root on both sides:

`(√(4x-4))^2\leq (4)^2`

`4x-4\leq 16`

`x\leq (20)/(4)`

`x\leq 5`

28.
`√(b+12) -√(b)>2`

Adding
`√(b)` to both the sides to get:

`√(b+12) -√(b)+√(b)>2+√(b)`

`√(b+12) >2+√(b)`

Taking square on both sides:

`(√(b+12))^2 >(2+√(b))^2`

`b+12>(2+√(b))^2`

`b+12>4+4√(b)+b`

`4+4√(b) +b<b+12`

Subtracting
`b` from both sides to get:

`4+4√(b) +b-b<b+12-b`

`4+4√(b) <12`

Subtracting 4 from both sides:

`4+4√(b)-4 <12-4`

`4√(b) <8`

Square both sides again:

`(4√(b))^2 <(8)^2`

`16b<8^2`

`b<(64)/(16)`

`b<4`

and for non-negative radical
`b\geq 0`

therefore, solution is
`0\leq b<4`.