Questions on Algebra: Radicals -- complicated equations involving roots answered by real tutors!

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Question 1095268: how much xsquare -y square ? if x+y=66 xy=9

Found 4 solutions by KMST, ikleyn, greenestamps, MathTherapy:
Answer by KMST(5345)   (Show Source):
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ONE WAY:
Maybe we know about solving quadratic equations.
Sometimes we can solve a quadratic equation such as by factoring if we find two integers such that and .
Then those values for and are the solutions of the equation and the equation is really
.
The solutions to a quadratic equation of the form are always two numbers whose sum and product can be found are and respectively.
Unfortunately, sometimes those numbers are irrational, or even imaginary, and then factoring is not an option.
Then, w must use algebra to "complete the square" and then solve, or apply the dreaded quadratic formula
that says the solutions to an equation of the form are given by

To avoid confusion, I used for the variable instead of , and I prefer equations where the leading coefficient is , so I wrote my equation as .
I will make and to get .
The solutions will be the numbers and (or and ) that add up to and whose product is .
The quadratic formula tells me that the solution are given by

The solutions to the equation are and
One is and the other is , but there is no way to guess which was intended to be and which was intended to be .
and are
and

The possible answers are
and

ANOTHER APPROACH:
We know that . If we only knew the value of we could easily find the value of

We know that , but to calculate we would need the value of

We know that and we know the values of and
Substituting the known values we get --> -->
Now we can use that value of to calculate the values of and , and from that find the value of

Then

Multiplying times

Answer by ikleyn(53748)   (Show Source):
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.
How much x square -y square ? if x+y=66 xy=9
~~~~~~~~~~~~~~~~~~~~~~~

Tutor @greenestamps provided the answer = .

This answer is correct, but incomplete.

This problem has a symmetry (x.y) <---> (y,x).

In other words, values x and y can be rearranged.

It means that together with the answer =
another answer = is valid and is possible, too.

So, the complete answer to the problem is THIS:

        - under given conditions, may have two different values.
           One possible value is ; another possible value is .

Answer by greenestamps(13327)   (Show Source):
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Given: x+y=66; xy=9

Find:

Answer by MathTherapy(10806)   (Show Source):
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Question 316943: simplify: sqrt (121) / (9)
a) sqrt (11) / sqrt (3)
b) 3 sqrt (11)
c) 11/3
d) sqrt (11) / (3)
Answer by MathTherapy(10806)   (Show Source):
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simplify: sqrt (121) / (9) a) sqrt (11) / sqrt (3) b) 3 sqrt (11) c) 11/3 d) sqrt (11) / (3) ****************** The other person's answer, "...d because the sqrt of 121 is 11 and the sqrt of 9 is 3," is PARTIALLY WRONG! If it's , then it's If it's , then it's

Question 1030250: I need help simplifying this expression:

I removed the common factor out of the square root to obtain , but the answer key says it is .
How is it possible? Am I missing out on a rule here?
Found 2 solutions by greenestamps, MathTherapy:
Answer by greenestamps(13327)   (Show Source):
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Consider this patern:

Given an expression to be simplified in the form , if you can modify the expression so the coefficient "b" on the inside radical is 2, then the expression can be simplified using that pattern.

In your example, you can take "4" out of the coefficient "8", leaving the required coefficient "2"; the "4" goes back inside the radical as :

The pattern now says you want to find integers a and b for which a+b=24 and ab=80. Those integers are 20 and 4, so

Answer by MathTherapy(10806)   (Show Source):
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I need help simplifying this expression: I removed the common factor out of the square root to obtain , but the answer key says it is . How is it possible? Am I missing out on a rule here? ************************************* This happens to be one of those RARE cases when you can actually factor out a COMMON factor, which happens to be a PERFECT SQUARE (in 4). This is what you did: = ----- Applying = ....At this point, you haven't fully simplified the expression, and should've continued as follows: ---- Changing 6 to 5 + 1, and 5 to 5*1 ---- Applying = --- Converting The above is in the form: , with , and so: then becomes: ----- Cancelling SQUARE and SQUARE ROOT = This certainly matches the answer key: ===== On the other hand, this author would've SIMPLIFIED the SURD from the onset, as follows: ----- Replacing 8 with factors, 2 & 4 ----- Converting 4 to ----- Applying = ---- Changing 24 to 20 + 4, and applying = ---- Converting The above is in the form: , with , and so: then becomes: ----- Cancelling SQUARE and SQUARE ROOT This certainly matches the answer key:

Question 973976: what is (4- square root of 8) divided by (2+ square root of 8) simplified

Answer by MathTherapy(10806)   (Show Source):
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what is (4- square root of 8) divided by (2+ square root of 8) simplified ********************************************* According to the other person, "I put this into the calculator at Mathway.com, and it comes out as: −4+3√2 (that's "three square root two")." Is this really "HELP?" Punching numbers on a calculator? Or, is this OUTRIGHT LAZINESS? * ---- Multiplying NUMERATOR and DENOMINATOR by , the DENOMINATOR'S conjugate ---- Converting to =

Question 78896: please help me simplify this problem?
sqrt/sqrt
Found 3 solutions by ikleyn, timofer, MathTherapy:
Answer by ikleyn(53748)   (Show Source):
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.
please help me simplify this problem?

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

The right way to simplify is THIS

         = = = .         ANSWER

                Solved completely and correctly.

According to commonly accepted agreements about square root,
the expression ' ' always goes from the square root as |a|, the absolute value of 'a'.

In this form, this transformation works universally for positive and negative values of ' a '.

The given problem is a standard TRAP to catch of those, who are unfamiliar with this trap for advance.

My post shows and teaches a standard way to avoid this trap.

As tutor @MathTherapy likes to say, " do not accept any other answer ".

Answer by timofer(155)   (Show Source):
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Answer by MathTherapy(10806)   (Show Source):
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please help me simplify this problem? sqrt/sqrt ************************************* The other person wrote, "Now, you need to see that , so you can also pull out a 4 from under the radical" How does this make sense? In what universe is ? As a result, his answer, is obviosly WRONG!! = = =

Question 669305: Please help me simplify this: sqrt(x^4y^7)
Answer by MathTherapy(10806)   (Show Source):
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Please help me simplify this: sqrt(x^4y^7) ************************************* The other person's answer, , is WRONG as he/she failed to SIMPLIFY this expression, completely. = =

Question 60849: solve for x by using the quadratic formula: 9x^2-6x+5=0
thanks so much
Answer by ikleyn(53748)   (Show Source):
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.
solve for x by using the quadratic formula: 9x^2-6x+5=0
thanks so much
~~~~~~~~~~~~~~~~~~~~~~~

        The solution by @jai_kos is incorrect.
        See my correct solution below.

9x^2-6x+5=0

It is the quadratic equation of the form.

ax^2+bx +c =0

Comparing the equations, we get
a = 9, b = -6 and c = 5

x =

x =

x =

x =

x = = .

Thus the given equation has no real solutions.
It has two conjugate solutions in complex numbers.

Solved correctly.

Question 60848: help on this please,,thanks
x+1= sqrt x+7
Answer by ikleyn(53748)   (Show Source):
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.
help on this please,, thanks
x+1= sqrt(x+7)
~~~~~~~~~~~~~~~~~~~~~~~~

        The solution in the post by @jai_kos is incorrect.
        See my correct solution below.

x+1= sqrt(x+7)

Square on both sides, we get

(x+1)^2 = x +7
x^2 + 2x + 1 = x + 7

Reduce it to the standard form quadratic equation

x^2 + x - 6 = 0

Factor left side

(x+3)*(x-2) = 0

The roots of this equation are x = -3 and x = 2.
Direct check shows that x = 2 is the solution to the original equation, while x = -3 is an EXTRANEOUS solution.

ANSWER. The given equation has a unique solution in real numbers x = 2.

Solved correctly.

Question 681800: I'm trying to prove that

Answer by MathTherapy(10806)   (Show Source):
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I'm trying to prove that ******************* This author's proof is different from the one the other person who responded, provided. His method may not be what the person who needs help, is looking for. Then again, this hunch may be WRONG! Having said that, let's focus on the L.H.S. because that's what's needed to be proven to equal 2. + + ---- Changing 6 to 2(3) + ------- Changing 3 to + ----- Changing 4 to 3 + 1, 3 to , and 12 to 9 + 3 + ------ Converting The above is in the form: + , with , and so: + then becomes: + + ----- Cancelling SQUARE and SQUARE ROOT + QED!

Question 1107093: One pair of integers (x.y) solves , find x*y.
Answer by MathTherapy(10806)   (Show Source):
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One pair of integers (x.y) solves , find x*y. *************************************************************** ----- Replacing 4 with its PRIME factors, 2 & 2 ----- Converting 2 to ---- Changing 11 to 7 + 4 ---- Converting The above is in the form: , with , and so: then becomes: ----- Cancelling SQUARE and SQUARE ROOT √(4√(7+11)) = , is in the form: y + √x, with y = 2, and x = 7. Therefore, x*y = 7(2) = 14.

Question 1124474: How would you simplify the equation:
Sqrt(17-12sqrt2) in the form
(a+bsqrt2)
Thank you.
Answer by MathTherapy(10806)   (Show Source):
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How would you simplify the equation: Sqrt(17-12sqrt2) in the form (a+bsqrt2) Thank you. ********************************** Not an equation, but an expression, instead!! ----- Changing 12 to 2*6 ------ Converting 6 to ------ Applying ----- Changing 17 to 9 + 8, and 72 (in ) to 9*8 ------ Applying ------ Converting The above is in the form: , with , and so: then becomes: ----- Cancelling SQUARE and SQUARE ROOT

Question 263259: solve: x^5/6 + x^2/3-2x^1/2 =0
thanks!
Answer by ikleyn(53748)   (Show Source):
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.
solve: x^5/6 + x^2/3 - 2x^1/2 = 0
~~~~~~~~~~~~~~~~~~~~~~~~~

        The solution in the post by @mananth is FATALLY incorrect.

        His first step is to raise all three addends in the left side of the equation to degree 6.
        But this step is not an equivalent transformation, so, starting from this point,
        his solution is inadequate.

        See my correct solution below.

The original equation is x^5/6 + x^2/3 - 2x^1/2 = 0. (1) It is the same as x^5/6 + x^4/6 - 2x^3/6 = 0. (2) Factor left side, taking x^3/6 out the parentheses as a common factor x^3/6*(x^2/6 + x^1/6 - 2) = 0. (3) One root is x = 0, generated by the common factor x^3/6. To find the roots generated by the expression in parentheses, introduce new variable t = x^1/6. Then the expression in parentheses is t^2 + r - 2. Its roots are = = = = . So, the roots are t = -2 and t = 1. Since t = x^1/6, we accept the positive root t = 1 and reject the negative root t = -2. Thus, finally we have the solutions to the original equation x = 0 and x = 1. ANSWER. The original equation has two real solutions x = 0 and x = 1.

Solved correctly.

Question 1094302: Solving by substitution with u

creates quadratic =

u = 2,-1
going back and using those two outputs in
and
and
Finally, I get and
But -2 can't be in a radical and would 4 turn into 2,-2 or just 2?
Answer by MathTherapy(10806)   (Show Source):
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Solving by substitution with u creates quadratic = u = 2,-1 going back and using those two outputs in and and Finally, I get and But -2 can't be in a radical and would 4 turn into 2,-2 or just 2? ****************************************************************** u = 2,-1 <=== This is okay! "going back and using those two outputs in and <=== Here's where I guess you got confused, and substituted 2 and - 1 for x in But, u = 2, as you mentioned above, NOT x = 2. And, because you'd substituted u for earlier, at this juncture, you need to BACK-SUBSTITUTE the value of u to get: ---- Squaring both sides (x - 4)(x + 1) = 0 x - 4 = 0 OR x + 1 = 0 x = 4 OR x = - 1 Now, you have 2 values for x that you can CHECK to ensure that they're VALID and NOT EXTRANEOUS. Also, u = - 1, as you mentioned above. And, because you'd substituted u for earlier, at this juncture, you need to BACK-SUBSTITUTE the value of u to get: Seeing that the square root of ANY expression is positive (> 0), it's obvious that u = - 1 is an EXTRANEOUS value. As such, x = 4, or x = - 1 (see above).

Question 569284: solve the following radical:
Answer by MathTherapy(10806)   (Show Source):
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solve the following radical: **********************************************< Just like "Rational-functions/875226", - 12 is NOT a solution. It's an EXTRANEOUS root. Only VALID and ACCEPTABLE answer:

Question 864099: Can you show me how to Rationalize the denominator and simply the equation
6 / the square root of 2 - the square root of 3.
Answer by MathTherapy(10806)   (Show Source):
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Can you show me how to Rationalize the denominator and simply the equation 6 / the square root of 2 - the square root of 3. ********************************************************************* = = =

Question 864097: Can you show me how to Rationalize the denominator and simply the equation the square root of 6 / the square root of 5 - the square root of 3.
Answer by MathTherapy(10806)   (Show Source):
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Can you show me how to Rationalize the denominator and simply the equation the square root of 6 / the square root of 5 - the square root of 3. ********************************************************************* = = =

Question 707329: How would you solve this equation?

Answer by MathTherapy(10806)   (Show Source):
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How would you solve this equation? ************************************** The other person's response, that "You get -10+6i or -10-6i (complex results)", is WRONG11 As the smaller radicand, 2x MUST be greater than or equal to 0, we get: . So, . This gives us: , with ---- Squaring both sides ----- Squaring both sides x - 8 = 0. <==== VALID, and ACCEPTABLE solution for x, in THIS case!!

Question 144143: Hello. Please help me solve this problem: I need to find the simplest radical form and the approximate answer of .

x = 0.8944
Answer by MathTherapy(10806)   (Show Source):
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Hello. Please help me solve this problem: I need to find the simplest radical form and the approximate answer of . x = 0.8944 ***************** ---- Multiplying by LCD, 2x ----- Squaring each side However, the negative x-value, is EXTRANEOUS, therefore leaving the sole VALID x-value, . Great job!! You got up to this point: , but needed to go a little further, by RATIONALIZING the DENOMINATOR, as demonstrated above. Your decimal approximation, x = 0.8944, is also CORRECT. I don't know why you sought help. You didn't need it! Again, great job!!

Question 798582: Simplify if necessary. Then rationalize the denominator.

Answer by MathTherapy(10806)   (Show Source):
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Simplify if necessary. Then rationalize the denominator. ************************** CORRECT answer: , or , and NOT as the other person states.

Question 459722: Simplify each expression by rationalizing the denominator.
3/sqrt(7)
2sqrt(2)/sqrt(5)
3sqrt(2)/sqrt(6)
2sqrt(5)/sqrt(12)
Answer by MathTherapy(10806)   (Show Source):
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Simplify each expression by rationalizing the denominator. 3/sqrt(7) 2sqrt(2)/sqrt(5) 3sqrt(2)/sqrt(6) 2sqrt(5)/sqrt(12) ****************** This problem "asks" to SIMPLIFY by "rationalizing the denominator," NOT providing a calculated value, as the other person did! , or , or , or

Question 904562:
Answer by MathTherapy(10806)   (Show Source):
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***************************** It is UNBELIEVABLE that the person who responded to this problem applied this RIDICULOUS approach. The person got up to , and UNBELIEVABLY, HE chose to, "complete the squares........" What would behoove someone to suggest such a thing to a person who asks for help with a math problem? Now, isn't SIMPLY x(x - 6) = 0, which results in values of 0 and 6 for x? Why would someone go through all that trouble to COMPLETE the SQUARE here? It's the same as using the quadratic equation formula to solve it, which is EXTREMELY RIDICULOUS and TOTALLY UNNECESSARY.....a TOTAL waste of time, in this author's opinion. Some people are probably more eccentric than is realized!!

Question 296793: Please help me solve this equation

Answer by MathTherapy(10806)   (Show Source):
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Please help me solve this equation ****************************** This is UTTERLY CRAZY!! Who ever heard of this: square everything <=== Who in their right mind would square both sides of a square root problem like this? Why do these people respond in this manner? This "HELP" is NO HELP at all!! Just leave the problem for someone who can handle it! BTW, the person claimed this to be an UNDEFINED equation. In other words, there are no solutions! Really!! Ha-ha-ha! --- Squaring BOTH sides, the CORRECT way ---- Dividing by GCF, 2 1 = 4(4 - x) --- Squaring each side 1 = 16 - 4x 1 - 16 = - 4x - 15 = - 4x This x-value, is VALID!!

Question 37745: Hi, I'm having trouble with this problem. Do you think you could help me?

Answer by MathTherapy(10806)   (Show Source):
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Hi, I'm having trouble with this problem. Do you think you could help me? *********************************************************************** Correct answer: , and NOT as the other person states!

Question 729072: square root of x+7=square root of 2x-3 (+2) not under the square root)
Answer by MathTherapy(10806)   (Show Source):
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square root of x+7=square root of 2x-3 (+2) not under the square root) ************************************************************************* I don't know what "X=507+338√2 ≈985.0042 from @lynnlo(4176) is all about. square root of x+7=square root of 2x-3 (+2) not under the square root) --- Squaring both sides ----- Squaring both sides (x - 42)(x - 2) = 0 x - 42 = 0 OR x - 2 = 0 x = 42 OR x = 2 However, x = 42 is an EXTRANEOUS solution, so ONLY solution is: x = 2

Question 519668: how do you solve the following:
+ =

Answer by MathTherapy(10806)   (Show Source):
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how do you solve the following: + = ****************************** One of the persons who responded states that , but this is WRONG!! x CANNOT equal . Of the 3 RADICAL expressions, 2x - 1 is the SMALLEST. So, , and ===> . So, we now get: + = , with + = = ----- Squaring both sides = 7x + 1 = 7x + 1 = 7x + 1 = 7x + 1 - 3x - 3 = 4x - 2 = ------ Squaring both sides 0 = (2x - 1)(x - 5) 0 = 2x - 1 OR 0 = x - 5 1 = 2x OR 5 = x As and , both solutions for x are VALID!

Question 701399: simplify
Answer by MathTherapy(10806)   (Show Source):
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simplify ==================================== The other person's answer, 15, is WRONG!! = = = = = = = = = =

Question 993742: Hello. I am having an issue solving this problem and I look to you for help. My problem is:
-+. By using prime factorization, I know that 12 = 2^2*3
50 = 5^2*2
72 = 3^2*2^3
Now, I plug these terms back in and I get:
-+ If I continue, I get:
-+
The answer key shows that the answer should be
- Can you see where I went wrong? I'm not coming up with the right answer.

Answer by MathTherapy(10806)   (Show Source):
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Hello. I am having an issue solving this problem and I look to you for help. My problem is: -+. By using prime factorization, I know that 12 = 2^2*3 50 = 5^2*2 72 = 3^2*2^3 Now, I plug these terms back in and I get: -+ If I continue, I get: -+ The answer key shows that the answer should be - Can you see where I went wrong? I'm not coming up with the right answer. ========================= - can NEVER be the simplified answer here. This answer must be for a different problem. Correct answer: ===================================================== Hello. I am having an issue solving this problem and I look to you for help. My problem is: -+. By using prime factorization, I know that 12 = 2^2*3 50 = 5^2*2 72 = 3^2*2^3 Now, I plug these terms back in and I get: -+ -+ If I continue, I get: -+ = +<==== Should be THIS, your FINAL answer!! You ALREADY have the correct answer, as you can see.

Question 1179144: For what values of c will x² +28x + c = 0 have no real solutions? Explain
Answer by ikleyn(53748)   (Show Source):
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.
For what values of c will x^2 +28x + c = 0 have no real solutions? Explain
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

The given quadratic equation has no real solutions if and only if its discriminant is a negative value. The discriminant is d = 28^2 - 4c = 784 - 4c. The condition of negativity the discriminant is 784 - 4c < 0, or 4c > 784, c > 784/4 = 196. ANSWER. The given equation has no real solutions at c > 196.

Solved.

Question 787079: Simplify the expression. Assume that all variables are positive
√x/5*√x/20
Answer by MathTherapy(10806)   (Show Source):
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Simplify the expression. Assume that all variables are positive √x/5*√x/20 If , then the answer is NOT 10x, as stated by the other person who responded. = = = =

Question 99475: Perform the indicated division. Rationalize the denominator if necessary. Then simplify each radical expression
-9-sqrt(108) dived by 3
Answer by MathTherapy(10806)   (Show Source):
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Perform the indicated division. Rationalize the denominator if necessary. Then simplify each radical expression -9-sqrt(108) dived by 3 The answer is (see Question # 83444), and NOT as the other person states. , as he/she suggests! I wonder where these people learned their mathematics!

Question 293344: how would you solve this proof?

Answer by MathTherapy(10806)   (Show Source):
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how would you solve this proof? This is indeed a NESTED SURD. From the other person's response, let me pick up from this point: Start with the given equation. Let's focus on the L.H.S. of the equation. --- Converting 2 to ---- Changing 8 to 6 + 2, and 12 to 6 * 2 --- Converting 6 to and 2 to The above is now in the form: , with a being , and b being So, now becomes: ---- Cancelling square and sqrt = As seen, L.H.S. = R.H.S.

Question 485164: Write the following expression as a radical and simplify if possible.
(-64)^4/3
Answer by MathTherapy(10806)   (Show Source):
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Write the following expression as a radical and simplify if possible. (-64)^4/3 The other person's answer, 81, is WRONG!! = =

Question 732218: how do you do √50x√18?
Found 2 solutions by MathTherapy, ikleyn:
Answer by MathTherapy(10806)   (Show Source):
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how do you do √50x√18? √50 x √18 = 5√2 x 3√2 = 5(3)(√2 x √2) = 15√2 x √2 = 15√4 = 15(2) = 30

Answer by ikleyn(53748)   (Show Source):
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.
how do you do √50x√18?
~~~~~~~~~~~~~~~~~~~~~~

I treat the square roots as having positive values and 'x' as the multiplication sign. It is what should be told in the problem's description, but since it is omitted, I insert it to make your request sensible. Then * = = = 30. ANSWER

Solved.

Question 735358: Rationalize 2 divided by 1 minus the square root of 2
Answer by ikleyn(53748)   (Show Source):
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.
Rationalize 2 divided by 1 minus the square root of 2.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

= . In the numerator, you have = . In the denominator, you have = 1 - 2 = -1. Therefore, the original fraction is . ANSWER

Solved.

Ignore the post by @lynnlo, since it is INCORRECT.

Question 736159: I am not sure how to work this problem out. (It asks for me to show my work)
x^5*x^√
Answer by ikleyn(53748)   (Show Source):
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.
I am not sure how to work this problem out. (It asks for me to show my work)
x^5*x^√
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

As the "problem" is written, it does not make sense.

So, this collection of words and mathematical symbols is NOT a Math problem
and, THEREFORE, is NOT a subject for discussion as a Math problem at this forum.

My condolences.

Question 633567: Please help me solve, a=?, b=35, c= 37
Found 3 solutions by mccravyedwin, Edwin McCravy, ikleyn:
Answer by mccravyedwin(421)   (Show Source):
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As always, Ikleyn has to be a smarty pants. She constantly criticizes the struggling students who post on this site. She mainly wants to show how clever she is, even though the clever method she uses is often not the one the teacher is teaching. So now she has to criticize me for criticizing her. Before she wrote this I was thinking of congratulating her for not criticizing this student for failing to give the necessary equation. No, I'm not here to "show my presence". It is important to show the student that if the question were this, Given , b=35, c=37, what is a? then the answer could be a = 12 or a = -12. Edwin

Answer by Edwin McCravy(20077)   (Show Source):
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12 or -12 Edwin

Answer by ikleyn(53748)   (Show Source):
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.
Please help me solve, a=?, b=35, c= 37
~~~~~~~~~~~~~~~~~~~~~~~~~

I consider it as a Pythagorean equation

a^2 + b^2 = c^2,

a^2 + 35^2 = 37^2,

a^2 = 37^2 - 35^2 = (37-35)*(37+35) = 2*72 = 144,

a^2 = 144,

a = = 12.

--------------------------------

Looking at Edwin's post, I clearly see that Edwin wants to show his presence here,
but I don't see much point in his introducing a negative value for the quantity
representing the length of a segment.

Only so I could refute it.