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So I'm going to assume that this question is asking for <u>non extraneous solutions</u>, or solutions that are found in the equation <em>and</em> are valid solutions when plugged back into the equation. So firstly, subtract 2 on both sides of the equation:
Next, square both sides:
Next, subtract x and add 2 to both sides of the equation:
Now we are going to be factoring by grouping to find the solution(s). Firstly, what two terms have a product of 6x^2 and a sum of -5x? That would be -3x and -2x. Replace -5x with -2x - 3x:
Next, factor x^2 - 2x and -3x + 6 separately. Make sure that they have the same quantity on the inside of the parentheses:
Now you can rewrite the equation as
Now, apply the Zero Product Property and solve for x as such:
Now, it may appear that the answer is C, however we need to plug the numbers back into the original equation to see if they are true as such:
Since both solutions hold true when x = 2 and x = 3, <u>your answer is C. x = 2 or x = 3.</u>
Given:
In a two-digit number, the tens digit is 5 less than the units digit.
The number itself is five more than three times the sum of its digits.
To find:
The number.
Solution:
Let the two digit number is ab. So,
Tens digit is 5 less than the units digit.
...(i)
The number itself is five more than three times the sum of its digits.
...(ii)
Using (i) and (ii), we get
Putting b=8 in (i), we get
Therefore, the required number is 38.
Answer:
- 15
Step-by-step explanation: