bash - Using grep or sed to substitute words not matching a pattern

I am new to the regular expression paradigm and I've encountered a problem I'm trying to solve, with no success.

Imagine a file test.txt with:

hello everyone, whatsi up
i hope my program worksa
if it doesnt... ho well!

I would like to output to another file output.txt only the words that start with a consonant and end with a vowel so it would result in:

hello whatsi
hope worksa

ho

I am using grep -o '\b[^ aeiouAEIOU]\w*[aeiouAEIOU]\b' test.txt > output.txt however, the -o flag outputs every matched string onto a new line. What should I do to get the format that I want? Another valid option would be to substitute everything that doesn't match that pattern with a blank space using sed but I failed to do it as well. Should I be using sed or awk instead?

Thanks

Answer

Following awk solution could help you in same too.

awk '{for(i=1;i<=NF;i++){if(tolower($i) ~ /^[^aeiou].*[aeiou]$/){val=val?val OFS $i:$i}};print val;val=""}'  Input_file

Output will be as follows.

hello whatsi
hope worksa
ho

Adding a non-one liner form with explanation too here.

awk '{
for(i=1;i<=NF;i++){       ##Starting a for loop here which starts from variable i value from 1 to till the value of NF(number of fields) value.
  if(tolower($i) ~ /^[^aeiou].*[aeiou]$/){ ##checking here condition if a field value in lower is satisfying the regex where I am checking if a value NOT starts from vowel and it is ending with vowels.
    val=val?val OFS $i:$i ##Creating a variable named val which will have value of current field value and it will concatenate its own value.
}
};
  print val;              ##Outside of loop, I am printing the value of variable val here, which will have all those words which are satisfying your conditions.
  val=""                  ##Nullifying the value of variable val here.
}

' Input_file              ##Mentioning the Input_file name here.