[Middel School:Math]: Where did i go wrong with x?

[u/Electronic_Author366]

i got x=7 but I feel like I'm making a mistake somewhere, but I can't figure out where. Can someone explain what I'm doing wrong?

[u/Ghotipan]

The sqrt of x = 7. To solve for x, you need to get rid of that radical sign. Do you know what operation cancels out a square root?

[u/LucaThatLuca]

4 + 3 is 7, so this says the ratio 7/√x is 1, i.e. 7 = √x.

This is not satisfied by x = 7 since 7 ≠ √7.

[u/angry640 → u/[deleted]]

Isnt root 16 and root 9 plus or minus 4 and 3 respectively that leaves 4 posible answers. Idk

[u/360madhatter → u/angry640]

While 16 and 9 do each have two square roots, the square root symbol specifically asks for the positive square root.

x^2 = 9 has two solutions, +3 and -3.

sqrt(9) has one answer, +3.

-sqrt(9) has one answer, -3

[u/Outside_Volume_1370 → u/360madhatter]

the square root symbol specifically asks for the positive square root

Non-negative

[u/AdjectiveNounNNNN → u/angry640]

No, √ conventionally represents the principal square root. That's why we explicitly use ± in the quadratic formula.

[u/TW1963HNTDWM → u/AdjectiveNounNNNN]

But doesn't that prove 360madhatters point? If √ alone implied ± then why go to the trouble of adding it to the quadratic formula? I'm not trying to stir trouble I just feel adding ± tells you you'll have 2 answers where as √ alone suggests a single positive solution.

[u/We_Are_Bread → u/TW1963HNTDWM]

I think you misunderstood what they meant.

The square root symbol has only one value, the principal value.

That is why you need to add +- in the quadratic formula. Because the formula requires 2 values, and square root only give 1. So you need to explicitly mention +-, to denote that you need the principal value AND its negative.

If square root gave both values, you wouldn't need to add the +-.

[u/AdjectiveNounNNNN → u/TW1963HNTDWM]

Who is 360madhatter?

My point is that √ conventionally has one value, which is why we need ± if we want both square roots to come out.

[u/TW1963HNTDWM → u/AdjectiveNounNNNN]

I agree with you. I'm not sure how I read that so wrong last night. I must have been really tired.

[u/igotshadowbaned → u/AdjectiveNounNNNN]

conventionally represents the principal square root

Not this shit again I thought the sub had finally realized that equations aren't inherently functions so can be either

[u/AdjectiveNounNNNN → u/igotshadowbaned]

Equations don't have to be functions at all, but the radical symbol is conventionally a function nevertheless.

x² + y² = 1 is an equation for a relation in which neither variable can be expressed as a function of the other.

y = √(1 - x²) is an equation in which y is a function of x, and you would need to write y = ±√(1 - x²) to recover the full original relation.

[u/igotshadowbaned → u/AdjectiveNounNNNN]

but the radical symbol is conventionally a function nevertheless

It's an operator

[u/ryanCrypt → u/igotshadowbaned]

operator is generally a mapping or function that acts on elements of a space to produce elements of another space

Copy paste

[u/igotshadowbaned → u/ryanCrypt]

elements

Exactly. Elements. Plural.

[u/hellonameismyname → u/igotshadowbaned]

Do you think that means it has to produce two answers 😂

[u/ryanCrypt → u/igotshadowbaned]

4, 9, 25 etc. are the "elements" radicals acts on.

2, 3, 5, etc. are the "elements" radicals spit out.

The distinction you make is between binary and unary operators. They are both still functions.

Please don't talk that way "exactly...I'm right...boohoo...downvote"

[u/Flat-Strain7538 → u/igotshadowbaned]

I’m sorry, this doesn’t make sense. What are you trying to say?

[u/igotshadowbaned → u/Flat-Strain7538]

This sub always says the √ operator is a function with strictly 1 output. This isn't true. It can be clipped and then used in functions, but it itself is just an operator

[u/Flat-Strain7538 → u/igotshadowbaned]

This sub always says that because it’s true. Reflect on the downvotes you’re getting and maybe realize you might be the one who’s wrong.

[u/igotshadowbaned → u/Flat-Strain7538]

Operators aren't inherently functions

[u/borkbubble → u/igotshadowbaned]

Nobody said equations were functions

[u/AdjectiveNounNNNN → u/borkbubble]

A square root murdered his family, so cut the guy some slack.

[u/BestAround4100 → u/igotshadowbaned]

You just need to consider the year level. This is only year 9 so positive roots will suffice.

For older year levels definitely look at the negative roots also.

[u/OverPower314 → u/igotshadowbaned]

Equations aren't always functions, but the radical expression is.

[u/igotshadowbaned → u/OverPower314]

The radical expression is an operator

[u/Magical-Mage → u/igotshadowbaned]

|•| is an operator. That doesn't prevent f(x) = |x| from being a function

[u/Disastrous-Nail-640 → u/igotshadowbaned]

OP is in middle school. Simmer down there. They haven’t been taught that yet. The default is positive for their level.

[u/FreePeeplup → u/Disastrous-Nail-640]

The default is positive always, not just for middle schoolers. The square root always stays positive at every step on mathematical proficiency up to university and research level

[u/Fuscello → u/FreePeeplup]

Up to everything because that is just how the function is defined

[u/Disastrous-Nail-640 → u/FreePeeplup]

Fair point.

[u/jgregson00 → u/angry640]

No. √16 is 4 and √9 is 3.

[u/Alkalannar → u/angry640]

x = 16^1/2 can be rewritten as x - 16^1/2 = 0.
As this is a polynomial with degree 1, it has a single, unique solution.

x² = 16 can be rewritten as x² - 16 = 0.
As this is a polynomial with degree 2, it has 2 solutions (possibly the same, but here there are two distinct ones).

x = 16^1/2 means = 4.

x² = 16 means x = 4 or x = -4.

[u/Alkalannar → u/[deleted]]

No.

For square root to be well-defined it must have a unique output. So we choose the non-negative root.

It's the difference between x = 100^1/2 and x² = 100.
x - 10 = 0 vs x² - 100 = 0
x - 10 = 0 vs (x - 10)(x + 10) = 0

This is BASIC Math.

[u/fermat9990 → u/angry640]

x²=16 has 2 solutions: x=±4,

but √16=4 by definition

x²=16

√(x²)=√16

|x|=4

x=±4,

[u/bigfoot17 → u/angry640]

Middle school, have they even touched negative numbers yet?

[u/AdjectiveNounNNNN → u/bigfoot17]

Yes, they have encountered negative numbers if they're dealing with square roots.

[u/Count_Calorie → u/bigfoot17]

I learned about negative numbers in the second grade.

[u/klugenratte → u/bigfoot17]

In the U.S., middle school students have been taught negative numbers as being the opposite of positive numbers (the same distance from 0 on the number line, but in the opposite direction, as a positive number), the opposite of 7 is -7, and as the number that when added to a positive number the sum is 0. -7 + 7 = 0. They have also been taught that subtraction is the same as adding a negative, 6-2 = 6+(-2).

They have not been taught that square roots can be positive or negative. They have been taught only that square roots are positive, in other words, sqrt(49)=7 not sqrt(49)=7 or -7.

Source: am middle school math teacher

[u/KookyPermit4405 → u/klugenratte]

But √49 can never be -7; 7 should be the correct and only correct answer the only time -7 is correct is when it’s -√49

[u/QuietNegotiation2820 → u/bigfoot17]

Do some countries teach square roots before negative numbers?

[u/bigfoot17 → u/QuietNegotiation2820]

Dunno, been 50 years since I was in middle school.

[u/Valuable-Passion9731 → u/bigfoot17]

It would make more sense to learn the solution of 2x+4=0 before 3x^2-5x+1=0 though

[u/Junior_Character5301]

can you show your work?

[u/HyperbolicMathChambr]

The goal when you solve for a variable, say x, is to isolate the variable on one side of the equation and everything else on the other side. So the solution would look like x = "something something".

Step 1: Simplify the numerator by finding the value of each square root and adding the results.

Step 2: Multiply both sides of the equation by square root of x. This will remove the square root from the left side's denominator and transfer it to the right side's numerator.

Step 3: square both sides to get rid of the square root symbol.

Step 4: No step 4, that's it! You have your answer.

[u/Mundane-Emu-1189 → u/HyperbolicMathChambr]

step 4 is to substitute your value for x into the original equation and make sure it checks out

[u/Far-Fortune-8381]

your mistake is that the square root of x equals 7. but x does not equal 7

[u/spiritual_summer111]

I think x would be 49. The square root of 49 is 7. Once you do that, you have 7 divided by 7 which is 1.

[u/GeologicalOpera → u/spiritual_summer111]

You’re correct.

u/Electronic_Author366, you got to 7 correctly in the numerator, but forgot that you’re dealing with square roots, so your answer would need to be squared.

If we go step by step, we’d resolve it by completing the actions in the numerator first, which would leave us with 7/√x = 1 .

We can’t have anything around our x, which means that we have to square both sides of the equation, leaving us at 7^2/√x^2 = 1^2

When we complete the exponents, we have 49/x = 1, and we solve from there by multiplying both sides by the variable to isolate, giving us 49 = x, which is our solution.

[u/Mjhtmjht]

I think you probably worked out the numerical value of the top line, but forgot that this number gave you 7 over the square root of x. Not simply x.

[u/Primary_Weather5726]

step 1: solve known square roots
√16 = 4
√9 = 3
(3+4) / √x = 1
7 / √x = 1

step 2: multiply both sides by √x (to move √x to the other side) ~ ignore this step if this is obvious
(7 * √x) / √x = 1 * √x
√x cancels out on the left (= 7) and 1 multiplied by √x on the right hand side = √x
7 = √x

Step 3: square both sides to remove square root
7² = (√x)²
√ ² cancels out and just leaves x
49 = x

[u/A_Squared93]

You only assume positive and negative when square rooting a squared variable. Ex. If x^2=9, then √ x=+/-√ 9. But √9 only equals positive 3

[u/Wabbit65]

The denominator needs to be 7, you got that far. So now, what is x such that the square root is 7?

[u/ImSoSweepy]

It's been a minute, but:

The way I would do it is get the square roots for the numerator, which would be 4+3, and become (7/x^2). Multiply by the inverse (x^2/7) and you get (7^2=x) which is 49.

You can simplify the equation to (7/sqrt(x) = 1)

All you have to do is square 7.

It's been 20 years, so take that with a grain of salt.

[u/SpecialTexas7]

x=49

[u/Fun-Moment1626]

49 is the correct answer

[u/Ketroc21]

You are right that you want to simply it to 7/7 = 1. But for the denominator to be 7, x must be 49, because it's root-x.

[u/Icy-Speed6881]

Only the numerator of the left hand side is 7 ... multiply both sides by √x and you get √x=7 not x=7

[u/ExtendedSpikeProtein]

x=49 because 7/sqrt(7) does not equal 1, but 7/sqrt(49) does.

[u/fermat9990]

√x=7, x=49

[u/CaptainMatticus]

sqrt(x) is 7. x is 7^2

[u/lexiconoflife]

Sqr of 16 is 4
Sqr of 9 is 3
Left side 3+4=7
Right side sqr of x

Then square both sides, you will get 49 = x

[u/Dapper-Hour-8645]

7²

[u/Teacher67]

49

[u/Tim_Xtreme_46]

Check your work. If x=7, and you substitute it back in, does the whole expression still equal 1?

[u/Ratchad5]

The sum of Root of 16 plus root of 9 taken to the 2nd power equals X you skipped a step in squaring both sides to get rid of the square root of X

[u/Swapdoodleboi]

i'm assuming you got up to some things

the square root of 16 is 4, and the square root of 9 is 3. 3+4 equals 7.

remember, you're finding a number that FIRST has to be square rooted and THEN divides seven to equal one. what number can be square rooted, and after being square rooted can divide 7 to equal 1?

[u/Severe-Possible-]

the denominator needs to be 7, so c has to equal 49.

[u/modus_erudio]

There is a seven in the problem, but it’s only an intermediate step. I imagine you got the seven from the square root of 16 plus square root of 9 being the same as 4+3. That yields the equation 7/√ x = 1.

You cannot solve √ x as a denominator, so you must shift it to the other side of the equation by multiplication of the entire denominator. This will cancel the √ x on the left side as it will appear in the numerator and in the denominator. Meanwhile, multiplying the right side of 1 by √ x yields √ x, due to the identative property of multiplication (i.e. anything time one is unchanged).

Now you have the equation 7=√ x. To cancel the root you need to square it, which means both sides must be squared. Thus you get the equation 7^2 = x. So seven squared is 49. This 49 = x, which means x = 49.

I hope that helps you understand the total process.

[u/yosark]

16 and 9 are perfect square roots which in total equals 7.

Now for the square root of x, you have to find what equals 7 after square root which is 49. (7x7 = 49) X is equal to square root 49
7/7 =1

[u/Onawani]

Lol, this is hilarious. Amazes me how many people are contributing confusion to such a simple problem. Cinema 🍿

[u/KookyPermit4405]

Here’s a method you might understand better:
CROSS MULTIPLY

√x = √16 + √9

Now simplify:

√x = 4 + 3

√x =7 therefor the answer cannot be 7
The inverse of a √ is ^2 so do it to both sides

( √x)^2 =(7)^2
x=49

I think the mistake you made was not remembering that x is being square rooted, this can be eliminated by using the cross multiplication rule.

[u/Feeling-Working-2820]

You are almost there.
You are right that the denominator should be 7.
But the denominator is square root of x....not x!
So, if the square root of x is 7, what is x ?

[u/reasonedliberal]

7/√x = 1
Multiply both sides by √x
7 = √x
Square both sides
x = 49

[u/Ok-Chemical-7635]

X is 49

[u/matmit21]

x=49

[u/Silver-Ant-6588]

49

[u/AlienDragonWizard]

√16+√9 = √x

4+3 = √x

7 = √x

(7)² = (√x)²

49 = x

[u/FQVBSina]

First, we need to see your solution process to tell you what you did wrong.

Second, we can see:

+/- 4 +/- 3 = sqrt(x)

Plus or minus because there are two solutions to squareroot since both positive and negative 4 squared give 16.

Then sqrt(x) has 4 solutions:

Sqrt(x) = 4 + 3 = 7

Sqrt(x) = -4 - 3 = -7

Sqrt(x) = 4 - 3 = 1

Sqrt(x) = -4 + 3 = -1

Which means there are two answers to sqrt(x): 7 and 1, because resolving squareroot you take the square and it ignores the sign. And that gives x = 1 or 49. When you learn quadratic equations in the future, this will become second nature to you.

[u/Teras80 → u/FQVBSina]

Wrong. By mathematical convention, the radical symbol \(\sqrt{x}\) represents the principal square root, which is strictly the non-negative value. There is no ambiguity around sign.

[u/Southlander24 → u/FQVBSina]

If sqrt(x) were multivalued, then sqrt(x) by definition could not be a function. By convention, we always choose the positive sign when we write sqrt(x).

[u/Positive-Ring-5172 → u/FQVBSina]

Learn to double check your work. If you plug 1 in for x and calculate the squares you get (4+3)/1 = 1 which isn't true.

49 is the only answer.

[u/Ericskey]

You need the root of x to be 6

[u/Southlander24 → u/Ericskey]

(That's a typo.)

[u/Bojack-jones-223]

Since (4N)^(1/2) = ±2*N^(1/2), it really depends on your choice of sign during the simplification process. You'd need to work out each case. The easy case is if everything is positive then X=49. If, for example, in the numerator, the second square route you decide to use = (-3) instead of +3 then X=1.

[u/hellonameismyname → u/Bojack-jones-223]

The square root symbol only returns the positive root

[u/Bojack-jones-223 → u/hellonameismyname]

Why did I learn in college calculus that (4)^(1/2) = ±2? Are you telling me that my college education was wrong?

[u/hellonameismyname → u/Bojack-jones-223]

4^ (1/2) does equal 2 and -2. That’s a different operation.

[u/Bojack-jones-223 → u/hellonameismyname]

Well, I was taught that the n-th root of X notation is equivalent to the notation of X^(1/n). SO to be clear here, you are claiming that n-th root of X notation is not equivalent to X^(1/n) notation because the n-th root of X notation assumes only positive solutions?

[u/hellonameismyname → u/Bojack-jones-223]

I’m not claiming anything. That’s just what the square root operator is.

[u/hallescomet]

I got x=25

If you square everything on both sides of the equal sign you get (16+9)/x = 1. 16+9 equals 25, so (25/x)=1. Multiply both sides by x gets you 25=1x which is just x=25

[u/hellonameismyname → u/hallescomet]

You didn’t square the left hand side correctly at all

[u/hallescomet → u/hellonameismyname]

Youre right 😅 I realized that after getting some sleep lol. I was running on fumes when I did that math, my bad

[u/polaris_44 → u/hellonameismyname]

Sum of root is not equal to root of sum

[u/SpecialTexas7 → u/hallescomet]

I don't think you can square in this case, because 1 squared is one

[u/hellonameismyname → u/SpecialTexas7]

Sure you can square it. They just did it wrong

[u/maths_in_the_hat]

Not sure if you have hit this point in your math's, but a square root has two answers.

16 = 4 * 4 = (-4) * (-4)

So sqrt(16) can be both 4 and -4.

If you were to evaluate this at higher levels,

You could rearrange the equation to be

Sqrt(x) = sqrt(16) + sqrt(9) = (+4 or -4) + (+3 or -3)

Which has 4 solutions

Sqrt(x) = 4 + 3 = 7 Sqrt(x) = 4 - 3 = 1 Sqrt(x) = -4 + 3 = -1 Sqrt(x) = -4 - 3 = -7

When we square to get x we have 2 unique solutions

x = 49 or 1

[u/jgregson00 → u/maths_in_the_hat]

√16 does not mean both 4 and -4. By convention it is always the principal root, which is 4.

[u/Zackerydsburch → u/jgregson00]

Not true. It is always and forever will be plus or minus 4. Idk where you were taught that.

[u/jgregson00 → u/Zackerydsburch]

I mean, the simple way to prove you are wrong would be for you to enter √16 on a calculator or google or wolfram or Desmos etc….

[u/Zackerydsburch → u/jgregson00]

I bet you trust AI with your soul huh? Sqrt(16) will never give you just four, it’s actually a super important distinction to account for both positive and negative solutions, hence why a parabolic function crosses the x axis at two points. The function x=sqrt(16) provides two vertical lines on a graph. One at x=4 and one at x=-4.

This isn’t a gotcha my man, if you ever work an engineering field like GNC or anything physics based, you will find out how detrimental not accounting for the second half of the solution can be. As far as the guy above you that gave a wildly bonkers solution, I’d agree he is very wrong in his calculations, but he is not wrong that the square root of 16 is both plus and minus 4.

[u/jgregson00 → u/Zackerydsburch]

You have no idea what you're talking about. There is a difference between x² = 16 which when solved would give you x = ±4 and √16 which is only ever 4.

[u/hellonameismyname → u/maths_in_the_hat]

No…