SOLUTION: Fill in the blanks to make an equation that is linear in disguise, and has a solution. sqrt(2x - 3) + __x + ___ = ___x - ___ + ____ 8 9 10 11 12 13 14 15 16 17 18 19 20

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Question 1210612: Fill in the blanks to make an equation that is linear in disguise, and has a solution.
sqrt(2x - 3) + __x + ___ = ___x - ___ + ____
8 9 10 11 12 13 14 15 16 17 18 19 20
Found 3 solutions by CPhill, KMST, greenestamps:
Answer by CPhill(2264)   (Show Source): You can put this solution on YOUR website!
To make the equation "linear in disguise," we need to eliminate the square root $\sqrt{2x - 3}$. The most common way to do this is to ensure the other terms allow you to isolate the root and then square both sides, or more simply, to choose coefficients that force the $x$ terms to cancel out in a way that leaves a solvable linear-like state.
However, the classic "in disguise" trick for this specific structure is to set it up so that when you isolate the radical and square it, the $x^2$ terms on both sides are identical, leaving you with a linear equation.
Here is a set of numbers from your list that works:
### The Equation
$$\sqrt{2x - 3} + \mathbf{8}x + \mathbf{11} = \mathbf{8}x + \mathbf{9} + \sqrt{\dots}$$
*(Note: Since you only provided one square root in the prompt structure, we will isolate it.)*
Let's fill the blanks to create a solvable linear path:
**$\sqrt{2x - 3} + 10x + 15 = 10x + 18$**
---
### Step-by-Step Solution:
1. **Subtract $10x$ from both sides:**
The $x$ terms cancel out immediately, which is the ultimate "disguise."
$$\sqrt{2x - 3} + 15 = 18$$
2. **Isolate the radical:**
Subtract $15$ from both sides:
$$\sqrt{2x - 3} = 3$$
3. **Square both sides:**
$$2x - 3 = 9$$
4. **Solve for $x$:**
$$2x = 12$$
$$x = 6$$
### Why these numbers?
* **10 (Blanks 1 & 3):** By putting the same coefficient for $x$ on both sides, the equation becomes a simple radical equation that reduces to linear form.
* **15 and 18 (Blanks 2 & 4):** We chose these so that $18 - 15 = 3$. When squared, this gives $9$, and since $9 + 3 = 12$ (an even number), it ensures $x$ is a clean integer ($6$).
**Filled Blanks:**
$\sqrt{2x - 3} + \underline{\mathbf{10}}x + \underline{\mathbf{15}} = \underline{\mathbf{10}}x + \underline{\mathbf{18}}$
*(The 5th blank in your prompt `+ ____` was not used in this specific solution, but could be filled with **0** or combined with the other constants).*
Would you like to try a version where the $x$ terms don't cancel immediately, requiring you to square a binomial?
Answer by KMST(5362)   (Show Source): You can put this solution on YOUR website!
The way I interpret the question, only the numbers given below the equation can be used to fill the blanks.
I am going to refer to the values to be filled as the constants A, B, C, D and E.
I will write the equation as
.
It is not specified, but I am going to assume that each number can only be used once.
A linear equation is one where the variable appears multiplied by some coefficient(s), but never with an exponent, or in a square root.
The square root shown appears to make that equation not linear, so
I am also going to assume, that the equation to be made is to be "linear in disguise" because
for the value of that satisfies .
<--> <-->
For to be a solution of , the values of the constants must be such that
<--><--><--> .
The values for A, B, C, D, and E must be all different, be in the list of numbers given, and the numbers given are such that .
The equation <--> tells us that
must be a multiple of 2, and
must be a multiple of 3.
From there on, we can guess and check to find suitable answers. There is a very large number of them.
For example, or give us
and are part of some of the many possible answers.
Those choices make ,
and leaves us the numbers 10, 11, 12, 13, 14, 15, 16, 17, 18, and 19 available to use as A and C.
Then, , , , , , , , could be used to complete an answer.
There are many other sets of values that are valid answers.

Let's verify if make an equation "linear in disguise" that has as a solution.
Substituting into , I get






For a more systematic approach, we could start with the possible values of .
Knowing that , and that is not zero,
the possible vales for are -12, -10, -8, -6, -4, -2, 2, 4, 6, 8, 10, and 12.
That makes the possible values for -18,-15,-12, -9, -6 -3, 3, 6, 9, 12, 15, and 18.
To get a longer list of possible solutions, we could start by using multiples of 3 for E, B, and D to make a multiple of 3.
Thinking about modular arithmetic, for those familiar with it, we could make a very long list of options for E, B, and D that make a multiple of 3
Then, we could weed out from those lists those with and those with , then we could find possible values of A and C, eliminating the ones where A or C was already used as E, B, or D.
We are still left with hundreds of different valid answers.
Answer by greenestamps(13351)   (Show Source): You can put this solution on YOUR website!


It is not at all clear what the purpose of this problem is. You can find numerous solutions either by pure trial and error or using formal algebra.

And, as another tutor pointed out, it is not clear whether numbers from the given list can be used more than once.

Using A, B, C, D, and E to represent the numbers in the blanks, the given equation is





This equation can be "linear in disguise" if x is any number for which is a perfect square.

Even limiting ourselves to integer perfect squares, there are an infinite number of possible values for x:

    2x-3   x
   ----------
     0    3/2
     1      2
     9      6
    25     14
    49     26
    ...

For many of those values of x the equation will have solutions (often multiple solutions) with A, B, C, D, and E being values from the given list.

But can be a perfect square which is not an integer; that will lead us to a whole new family of possible values for x, and possibly many of those will allow additional solutions to the problem.

Finally does not have to be a perfect square. We can isolate the radical and square both sides of the equation to get yet another whole family of possible solutions to the problem.




Squaring both sides will yield a quadratic equation; since we are supposed to get a linear equation, it would be necessary for to be zero -- i.e., A and C would have to be the same value from the given list. If that is indeed allowed, then we open up the possibility of many more solutions to the problem.
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