How do I simplify the square root of 4a to the 4th power, and the square root of 27ab to the 3rd power?
Q. I need help with the full specific description of the following equations: The square root of 4a to the 4th power, or 4a(superscript)4. The square root of 27ab to the 3rd power, or 27ab(superscript)3. The square root of 27ab to the 3rd power, or 27ab3. Sorry, double post of the second equation. Ignore that.
Asked by Ryan G. - Fri May 8 11:06:00 2009 - - 1 Answers - 0 Comments
A. Not clear if you mean: sqrt((4a)^4) or sqrt(4(a^4)) Same for other: sqrt((27ab)^3) or sqrt(27a(b^3)) Subtle I know, but makes a difference. Let me know and I'll check back.
Answered by tominator - Fri May 8 12:23:35 2009
Q. I need help with the full specific description of the following equations: The square root of 4a to the 4th power, or 4a(superscript)4. The square root of 27ab to the 3rd power, or 27ab(superscript)3. The square root of 27ab to the 3rd power, or 27ab3. Sorry, double post of the second equation. Ignore that.
Asked by Ryan G. - Fri May 8 11:06:00 2009 - - 1 Answers - 0 Comments
A. Not clear if you mean: sqrt((4a)^4) or sqrt(4(a^4)) Same for other: sqrt((27ab)^3) or sqrt(27a(b^3)) Subtle I know, but makes a difference. Let me know and I'll check back.
Answered by tominator - Fri May 8 12:23:35 2009
How do I figure out a square root squared?
Q. I am trying to determine if this is a right triangle. The legs are square root 10 and square root 5. The hypotenuse is square root 13. When I use the Pythagorean Theorem then my equation looks like this. square root 10 squared + square root 5 squared = square root 13 squared. Is this right? If it is how do I square a square root? Or am I making this harder than it really is?
Asked by Jen - Sun Sep 23 00:15:04 2007 - - 5 Answers - 0 Comments
A. Yes, you are making it hard. Square root cancels squares and vice-versa. Your next step is 10 + 5 = 13. So no this is not a right triangle.
Answered by faithlocket - Sun Sep 23 00:25:16 2007
Q. I am trying to determine if this is a right triangle. The legs are square root 10 and square root 5. The hypotenuse is square root 13. When I use the Pythagorean Theorem then my equation looks like this. square root 10 squared + square root 5 squared = square root 13 squared. Is this right? If it is how do I square a square root? Or am I making this harder than it really is?
Asked by Jen - Sun Sep 23 00:15:04 2007 - - 5 Answers - 0 Comments
A. Yes, you are making it hard. Square root cancels squares and vice-versa. Your next step is 10 + 5 = 13. So no this is not a right triangle.
Answered by faithlocket - Sun Sep 23 00:25:16 2007
How do I rationalize the denominator of the square root of 2 over the square root of 10?
Q. I keep getting the square root of 20 over the square root of 100, but I'm afraid that that will turn out as the wrong answer. I need help with rationalizing the denominator of the square root of 2 over the square root of 10, and I need a specific answer. Can someone please help me?
Asked by Ryan G. - Thu May 7 23:50:56 2009 - - 2 Answers - 0 Comments
A. Multiply the top and bottom by sqrt(10) so that you have sqrt(20)/sqrt(100) The square root of 100 is 10, so you have sqrt(20)/10. 20= 5*4 = 5*2*2, so sqrt(20) = 2*sqrt(5) 2*sqrt(5)/10 = sqrt(5)/5, or square root of 5 over 5.
Answered by unknown - Thu May 7 23:58:40 2009
Q. I keep getting the square root of 20 over the square root of 100, but I'm afraid that that will turn out as the wrong answer. I need help with rationalizing the denominator of the square root of 2 over the square root of 10, and I need a specific answer. Can someone please help me?
Asked by Ryan G. - Thu May 7 23:50:56 2009 - - 2 Answers - 0 Comments
A. Multiply the top and bottom by sqrt(10) so that you have sqrt(20)/sqrt(100) The square root of 100 is 10, so you have sqrt(20)/10. 20= 5*4 = 5*2*2, so sqrt(20) = 2*sqrt(5) 2*sqrt(5)/10 = sqrt(5)/5, or square root of 5 over 5.
Answered by unknown - Thu May 7 23:58:40 2009
How do work out the square root of a number in programming language?
Q. Im trying to write a function in Haskell to return the square root How can i do that without using a pre written function
Asked by Nicknam3 - Mon Sep 28 12:49:06 2009 - - 2 Answers - 0 Comments
A. I believe there is already a function in Haskell to calculate the square root (sqrt or isqrt, if you need complex results). If you wanted to calculate this, you would have to use successive approximations or a Taylor expansion.
Answered by KevinM - Mon Sep 28 13:08:36 2009
Q. Im trying to write a function in Haskell to return the square root How can i do that without using a pre written function
Asked by Nicknam3 - Mon Sep 28 12:49:06 2009 - - 2 Answers - 0 Comments
A. I believe there is already a function in Haskell to calculate the square root (sqrt or isqrt, if you need complex results). If you wanted to calculate this, you would have to use successive approximations or a Taylor expansion.
Answered by KevinM - Mon Sep 28 13:08:36 2009
What is the square root of the imaginary number, i?
Q. I know the imaginary number, i, is the square root of -1. But what about the square root of i, in other words, the square root of the square root of -1. Can you simplify the final value? I did some calculation on my own and found that it's still i. But I'm probably wrong. Any idea? Thx What about the square root of negative i? What would you get from that?
Asked by Minitrue - Mon Jan 7 22:23:54 2008 - - 6 Answers - 0 Comments
A. i = (1 + i)/ 2 (-i) = (1 - i)/ 2 You can find these values using DeMoivre's theorem: (cos + i sin )^n = cos (n ) + i sin (n ). Now, representing i in polar form, we have i = cos ( /2) + i sin ( /2). Whence it follows that i^(1/2) = (cos ( /2) + i sin ( /2))^(1/2) = cos ( /4) + i sin ( /4) = 1/ 2 + i/ 2, as above. Similar calculations yield the square root of -i.
Answered by Pascal - Mon Jan 7 22:39:27 2008
Q. I know the imaginary number, i, is the square root of -1. But what about the square root of i, in other words, the square root of the square root of -1. Can you simplify the final value? I did some calculation on my own and found that it's still i. But I'm probably wrong. Any idea? Thx What about the square root of negative i? What would you get from that?
Asked by Minitrue - Mon Jan 7 22:23:54 2008 - - 6 Answers - 0 Comments
A. i = (1 + i)/ 2 (-i) = (1 - i)/ 2 You can find these values using DeMoivre's theorem: (cos + i sin )^n = cos (n ) + i sin (n ). Now, representing i in polar form, we have i = cos ( /2) + i sin ( /2). Whence it follows that i^(1/2) = (cos ( /2) + i sin ( /2))^(1/2) = cos ( /4) + i sin ( /4) = 1/ 2 + i/ 2, as above. Similar calculations yield the square root of -i.
Answered by Pascal - Mon Jan 7 22:39:27 2008
How does a calculator calculate the square root of nonsquare numbers?
Q. Does it guess up and down continuously until it gets close enough?(because this could take some time so what does it do? Adding and subtracting would be fairly easy to do dividing is also straight forward as is multiplying but what about finding the square root? Can you give a little detail?
Asked by mete - Sat Sep 15 15:38:46 2007 - - 3 Answers - 1 Comments
A. There is a build-in procedure, involving just addition, subtraction, multiplication and division based on the following: take an arbitrary a > 0 and consider the recurrence relationship x_{n+1} = (1/2)(x_{n} + a/x_{n]), where n = 1,2,3,...; x_{1} = 1. Starting with x_{1} we can compute x_{2} by the above formula, then x_{3} etc. It is proven long ago that the sequence: x_{1}, x_{2}, x_{3}, . . . , x_{n}, . . is convergent and its limit is exactly sqrt(a) /You can try to prove that, it's a popular exercise/. More, the sequence is very convenient to calculate, only the 4 arithmetic operations involved, and the most important - it converges very quickly, what means that You need a few terms to be close to the limit. Normally the iteration… [cont.]
Answered by Duke - Sat Sep 15 16:05:24 2007
Q. Does it guess up and down continuously until it gets close enough?(because this could take some time so what does it do? Adding and subtracting would be fairly easy to do dividing is also straight forward as is multiplying but what about finding the square root? Can you give a little detail?
Asked by mete - Sat Sep 15 15:38:46 2007 - - 3 Answers - 1 Comments
A. There is a build-in procedure, involving just addition, subtraction, multiplication and division based on the following: take an arbitrary a > 0 and consider the recurrence relationship x_{n+1} = (1/2)(x_{n} + a/x_{n]), where n = 1,2,3,...; x_{1} = 1. Starting with x_{1} we can compute x_{2} by the above formula, then x_{3} etc. It is proven long ago that the sequence: x_{1}, x_{2}, x_{3}, . . . , x_{n}, . . is convergent and its limit is exactly sqrt(a) /You can try to prove that, it's a popular exercise/. More, the sequence is very convenient to calculate, only the 4 arithmetic operations involved, and the most important - it converges very quickly, what means that You need a few terms to be close to the limit. Normally the iteration… [cont.]
Answered by Duke - Sat Sep 15 16:05:24 2007
What is answer to the Log square root of 10 to the 3rd power?
Q. Its hard to write it out on the computer but it looks something a long the lines of: log power of three next to the square root of 10? Help?
Asked by V - Tue May 26 19:41:37 2009 - - 1 Answers - 0 Comments
A. // log(sqrt10)^3 =(3/2)log10=3/2
Answered by vect - Thu May 28 22:14:40 2009
Q. Its hard to write it out on the computer but it looks something a long the lines of: log power of three next to the square root of 10? Help?
Asked by V - Tue May 26 19:41:37 2009 - - 1 Answers - 0 Comments
A. // log(sqrt10)^3 =(3/2)log10=3/2
Answered by vect - Thu May 28 22:14:40 2009
Can the square root of any positive integer that is not a perfect square be irrational?
Q. For example, square root of 3 and the square root of 5 yields an irrational number. If not, how would you prove that all numbers that are not perfect squares be irrational?
Asked by jtrn69 - Wed Jul 18 16:02:37 2007 - - 2 Answers - 1 Comments
A. You mean, can it be RATIONAL? You could take the classic proof of 2's irrationality and expand it to this general case. Or start off with a fraction that is defined as not being an integer, and show that its square can't be an integer. Assume that there is an integer whose square root is not an integer nor an irrational number. Therefore it has to be a non-integer, rational number. Call this root a/b (where "a" and "b" are coprime integers, and b is not "1"). This means the original integer is (a/b)^2, or (a^2) / (b^2). If this is to be a fraction, then a^2 has to be a multiple of b^2. This means all of the prime factors of b^2 must also be prime factors of a^2, which in turn means that all of the prime factors of b must be prime… [cont.]
Answered by Geezah - Wed Jul 18 16:08:47 2007
Q. For example, square root of 3 and the square root of 5 yields an irrational number. If not, how would you prove that all numbers that are not perfect squares be irrational?
Asked by jtrn69 - Wed Jul 18 16:02:37 2007 - - 2 Answers - 1 Comments
A. You mean, can it be RATIONAL? You could take the classic proof of 2's irrationality and expand it to this general case. Or start off with a fraction that is defined as not being an integer, and show that its square can't be an integer. Assume that there is an integer whose square root is not an integer nor an irrational number. Therefore it has to be a non-integer, rational number. Call this root a/b (where "a" and "b" are coprime integers, and b is not "1"). This means the original integer is (a/b)^2, or (a^2) / (b^2). If this is to be a fraction, then a^2 has to be a multiple of b^2. This means all of the prime factors of b^2 must also be prime factors of a^2, which in turn means that all of the prime factors of b must be prime… [cont.]
Answered by Geezah - Wed Jul 18 16:08:47 2007
How can I prove that the square root of 8 minus the square root of 3 is an irrational number.?
Q. I am supposed to use thefact that the square root of 3 is an irrational number. I know I need to use a proof by contradiction, but I am stuck after that.
Asked by Ihave Y - Fri Nov 16 14:09:59 2007 - - 4 Answers - 0 Comments
A. Suppose 8 - 3 is rational. Then its square is rational, so we have that ( 8 - 3) = 8 - 2 24 + 3 is rational. Then subtracting 11 from it will also yield a rational number, so we have that -2 24 is rational. Since dividing a rational number by -2 yields a rational number, we have that 24 is rational. However, 24 is not a perfect square, so its square root cannot be rational -- a contradiction. Therefore, 8 - 3 is not rational.
Answered by Pascal - Fri Nov 16 14:30:34 2007
Q. I am supposed to use thefact that the square root of 3 is an irrational number. I know I need to use a proof by contradiction, but I am stuck after that.
Asked by Ihave Y - Fri Nov 16 14:09:59 2007 - - 4 Answers - 0 Comments
A. Suppose 8 - 3 is rational. Then its square is rational, so we have that ( 8 - 3) = 8 - 2 24 + 3 is rational. Then subtracting 11 from it will also yield a rational number, so we have that -2 24 is rational. Since dividing a rational number by -2 yields a rational number, we have that 24 is rational. However, 24 is not a perfect square, so its square root cannot be rational -- a contradiction. Therefore, 8 - 3 is not rational.
Answered by Pascal - Fri Nov 16 14:30:34 2007
Is the square root of a function that is not analytic necessarily not analytic in the complex plane?
Q. I'm trying to prove that the norm of z is nowhere differentiable using the definition of the limit in the complex plane. My thought is to square it, prove that the resulting function is not analytic, and taking the square root. Other methods/thoughts would be appreciated.
Asked by v - Mon Feb 9 20:13:12 2009 - - 1 Answers - 0 Comments
A. If I understand you correctly, your strategy is as follows: (1) Show that |z|^2 is nowhere differentiable. (2) Argue that because |z|^2 is nowhere differentiable, |z| can't be either. This will almost work, but there are two issues. The idea you're using in (2) is formulated more clearly without square roots. Textbooks often include a proof that if f and g are differentiable at a point z, then their product fg is also differentiable at z. In particular, if we take f and g to be the same: If f is differentiable at a point z, then so is f^2. This implies that: If f^2 is not differentiable at a point z, then f is not differentiable at z either. The reason for phrasing things in terms of squaring instead of square rooting is that… [cont.]
Answered by mcbengt - Wed Feb 11 23:07:16 2009
Q. I'm trying to prove that the norm of z is nowhere differentiable using the definition of the limit in the complex plane. My thought is to square it, prove that the resulting function is not analytic, and taking the square root. Other methods/thoughts would be appreciated.
Asked by v - Mon Feb 9 20:13:12 2009 - - 1 Answers - 0 Comments
A. If I understand you correctly, your strategy is as follows: (1) Show that |z|^2 is nowhere differentiable. (2) Argue that because |z|^2 is nowhere differentiable, |z| can't be either. This will almost work, but there are two issues. The idea you're using in (2) is formulated more clearly without square roots. Textbooks often include a proof that if f and g are differentiable at a point z, then their product fg is also differentiable at z. In particular, if we take f and g to be the same: If f is differentiable at a point z, then so is f^2. This implies that: If f^2 is not differentiable at a point z, then f is not differentiable at z either. The reason for phrasing things in terms of squaring instead of square rooting is that… [cont.]
Answered by mcbengt - Wed Feb 11 23:07:16 2009
How do you find the fourth root of a square root?
Q. Question: Log base 2 of the fourth root of the square root of 2? Anyone have an idea as to how to begin to simplify this problem.
Asked by hp - Mon Sep 14 23:50:13 2009 - - 1 Answers - 0 Comments
A. "fourth root" means an exponent of 1/4. "square root" means an exponent of 1/2. "fourth root of the square root" means an exponent of (1/4)*(1/2)=1/8 So, we are taking the log base 2 of 2^(1/8) Since our log base and our exponent base are the same, the log is the exponent, 1/8.
Answered by unknown - Mon Sep 14 23:58:02 2009
Q. Question: Log base 2 of the fourth root of the square root of 2? Anyone have an idea as to how to begin to simplify this problem.
Asked by hp - Mon Sep 14 23:50:13 2009 - - 1 Answers - 0 Comments
A. "fourth root" means an exponent of 1/4. "square root" means an exponent of 1/2. "fourth root of the square root" means an exponent of (1/4)*(1/2)=1/8 So, we are taking the log base 2 of 2^(1/8) Since our log base and our exponent base are the same, the log is the exponent, 1/8.
Answered by unknown - Mon Sep 14 23:58:02 2009
Why does the square root of 28x cubed equal the square root of 7x?
Q. I need someone to explain to me why the square root of twenty eight x cubed ( 28x^3) equals the square root of seven x ( 7x). I am completely stumped!
Asked by Jeff - Fri Jun 26 14:21:40 2009 - - 5 Answers - 0 Comments
A. It is NOT, it is a typo check your source. MoonRose9 is correct. (28x^3) = 2x (7x) after simplification.
Answered by Daoist - Fri Jun 26 14:46:48 2009
Q. I need someone to explain to me why the square root of twenty eight x cubed ( 28x^3) equals the square root of seven x ( 7x). I am completely stumped!
Asked by Jeff - Fri Jun 26 14:21:40 2009 - - 5 Answers - 0 Comments
A. It is NOT, it is a typo check your source. MoonRose9 is correct. (28x^3) = 2x (7x) after simplification.
Answered by Daoist - Fri Jun 26 14:46:48 2009
When six squared is divided by the square root of 81, what is the quotient?
Q. When six squared is divided by the square root of 81, what is the quotient?
Asked by guest - Sat Nov 15 14:08:02 2008 - - 2 Answers - 0 Comments
A. There are two answers: +4 and -4. 6 squared is 36. the square root of 81 is +9 and -9. 36 divided by +9 is +4, 36 divided by -9 is -4.
Answered by Joe H - Sat Nov 15 14:20:01 2008
Q. When six squared is divided by the square root of 81, what is the quotient?
Asked by guest - Sat Nov 15 14:08:02 2008 - - 2 Answers - 0 Comments
A. There are two answers: +4 and -4. 6 squared is 36. the square root of 81 is +9 and -9. 36 divided by +9 is +4, 36 divided by -9 is -4.
Answered by Joe H - Sat Nov 15 14:20:01 2008
What is the antiderivative of square root x times sin x?
Q. I spent an hour and I still can't figure this out. What is the anti-derivative of: square root (x) times sin(x)? Thanks.
Asked by sabershadezero - Tue Mar 24 20:45:36 2009 - - 2 Answers - 0 Comments
A. This is a nonelementary integral; therefore it can not be solved traditionally. Solution:
Answered by James - Tue Mar 24 21:02:50 2009
Q. I spent an hour and I still can't figure this out. What is the anti-derivative of: square root (x) times sin(x)? Thanks.
Asked by sabershadezero - Tue Mar 24 20:45:36 2009 - - 2 Answers - 0 Comments
A. This is a nonelementary integral; therefore it can not be solved traditionally. Solution:
Answered by James - Tue Mar 24 21:02:50 2009
How do you graph a square root on a number line?
Q. Homework help! =D How do you graph the square root of a number on a number line without using a calculator? Are there any special tricks or shortcuts you learned?
Asked by SensiblyDaft - Tue Aug 11 17:06:57 2009 - - 1 Answers - 0 Comments
A. I assume you mean one that is not rational, such as 2. In such a case, you make a reasonable estimate of it's position, and then label the point that you plot. For example, you know that 2 is greater than 1 and less than 2, so put the point at about 1 (actual value is about 1.4142). For 3, you know the answer is still less than 4, but greater than 2. If both of those points are required to be plotted just make sure you put it in proper relation, otherwise about 1 is plenty good (actual value is about 1.7321). If you are going to get into larger numbers, it's not a bad idea to just learn a few roots. Certainly 2, 3, and 5 (2.2361) and 10 (3.1623) shouldn't be too hard. Then for a number like 20, which you can quickly workout is 4 5… [cont.]
Answered by I'm with Stupid - Sat Aug 15 02:37:32 2009
Q. Homework help! =D How do you graph the square root of a number on a number line without using a calculator? Are there any special tricks or shortcuts you learned?
Asked by SensiblyDaft - Tue Aug 11 17:06:57 2009 - - 1 Answers - 0 Comments
A. I assume you mean one that is not rational, such as 2. In such a case, you make a reasonable estimate of it's position, and then label the point that you plot. For example, you know that 2 is greater than 1 and less than 2, so put the point at about 1 (actual value is about 1.4142). For 3, you know the answer is still less than 4, but greater than 2. If both of those points are required to be plotted just make sure you put it in proper relation, otherwise about 1 is plenty good (actual value is about 1.7321). If you are going to get into larger numbers, it's not a bad idea to just learn a few roots. Certainly 2, 3, and 5 (2.2361) and 10 (3.1623) shouldn't be too hard. Then for a number like 20, which you can quickly workout is 4 5… [cont.]
Answered by I'm with Stupid - Sat Aug 15 02:37:32 2009
How can I simplify square root of 735 divided by square root of 5 into the form A square root of B?
Q. I already know that I have to rationalize the denomenator which is multiplying square root of 5 to the numerator and denomenator but, i still get he wrong answer.
Asked by Ren - Sun Jun 1 21:25:50 2008 - - 5 Answers - 0 Comments
A. sqrt(735)/sqrt(5) = sqrt(735/5) = sqrt(147) = sqrt(3*49) = 7sqrt(3) this is it, here A = 7 and b = 3 hope this helps
Answered by Niecis aw Ignacy Ska ecki - Sun Jun 1 21:30:06 2008
Q. I already know that I have to rationalize the denomenator which is multiplying square root of 5 to the numerator and denomenator but, i still get he wrong answer.
Asked by Ren - Sun Jun 1 21:25:50 2008 - - 5 Answers - 0 Comments
A. sqrt(735)/sqrt(5) = sqrt(735/5) = sqrt(147) = sqrt(3*49) = 7sqrt(3) this is it, here A = 7 and b = 3 hope this helps
Answered by Niecis aw Ignacy Ska ecki - Sun Jun 1 21:30:06 2008
How do I recognize a square root function?
Q. How would I recognize a square root function in a table, graph, or an equation?
Asked by epor1125 - Wed Apr 22 18:51:21 2009 - - 1 Answers - 0 Comments
A. If you look at the graph, you would see something that looks like half of a sideways-U (or half of a parabola turned on its side). If you look at the equation, you will either see a radical, or the exponent will be 1/2 or 0.5. If you look at the table, you should notice that when you input perfect squares(for example 49, or 64), the results are usually whole numbers.
Answered by Damasta AM inductee - Wed Apr 22 19:06:42 2009
Q. How would I recognize a square root function in a table, graph, or an equation?
Asked by epor1125 - Wed Apr 22 18:51:21 2009 - - 1 Answers - 0 Comments
A. If you look at the graph, you would see something that looks like half of a sideways-U (or half of a parabola turned on its side). If you look at the equation, you will either see a radical, or the exponent will be 1/2 or 0.5. If you look at the table, you should notice that when you input perfect squares(for example 49, or 64), the results are usually whole numbers.
Answered by Damasta AM inductee - Wed Apr 22 19:06:42 2009
How do I write the square root of 24 in simplified radical form?
Q. How do you write the square root of 24 in simplified radical form?
Asked by Jem - Fri Oct 9 16:54:44 2009 - - 6 Answers - 0 Comments
A. sq rt 24 = sq rt (2 * 2 * 2 * 3) Now the inside of the square root is equal to 24. but you can take out the 2 :) so ... 2 sq rt (3 * 2) then 2 sq rt (6) is the answer! P:)
Answered by Dee - Fri Oct 9 17:02:19 2009
Q. How do you write the square root of 24 in simplified radical form?
Asked by Jem - Fri Oct 9 16:54:44 2009 - - 6 Answers - 0 Comments
A. sq rt 24 = sq rt (2 * 2 * 2 * 3) Now the inside of the square root is equal to 24. but you can take out the 2 :) so ... 2 sq rt (3 * 2) then 2 sq rt (6) is the answer! P:)
Answered by Dee - Fri Oct 9 17:02:19 2009
How do I make a square root sign on the keyboard?
Q. I'm suing an online graphing calculator and need to make a square root sign. Please help!
Asked by Abbe - Wed May 2 20:03:07 2007 - - 5 Answers - 0 Comments
A. You can get the square root sign with the keyboard shortcut: Alt+251, ie: hold down alt, press (on the numpad) 2, then 5, then 1, then release alt. This produces ' ' . But note, it is unlikely that an online graphic calculator uses this. It might just be sqrt(...) And if that does not work, square rooting is the same as raising to the 1/2 power, thus: ^(1/2) or ^.5 should work.
Answered by NSurveyor - Wed May 2 20:08:42 2007
Q. I'm suing an online graphing calculator and need to make a square root sign. Please help!
Asked by Abbe - Wed May 2 20:03:07 2007 - - 5 Answers - 0 Comments
A. You can get the square root sign with the keyboard shortcut: Alt+251, ie: hold down alt, press (on the numpad) 2, then 5, then 1, then release alt. This produces ' ' . But note, it is unlikely that an online graphic calculator uses this. It might just be sqrt(...) And if that does not work, square rooting is the same as raising to the 1/2 power, thus: ^(1/2) or ^.5 should work.
Answered by NSurveyor - Wed May 2 20:08:42 2007
What's the difference of solving a quadratic equation using square root and completing the square?
Q. I have 2 equations that I need to solve using the square root property. 2x^2-5=93 and (x+4)^2=81 and I need to explain how I got the answers. How would I do this?
Asked by Creole - Sat Aug 25 23:37:30 2007 - - 2 Answers - 0 Comments
A. To solve a equation using the method of 'square root' in a quadratic equation, the equation must be of the form (x + h)^2 = k. If the equation is not of the form (x + h)^2 = k, you would have to apply 'completing the square' method to manipulate a quadratic equation of the form ax^2 + bx +c = 0 to (x + h)^2 = k. 2x^2 - 5 = 93 2x^2 = 98 x^2 = 49 x = 7 or x = -7 (x + 4)^2 = 81 x + 4 = 9 or x + 4 = -9 x = 5 or x = -13 For these two equations, you can use the 'square root' method straightaway without applying 'completing the square'. Here is an example of a quadratic equation which requires you to 'complete the square' before you can use the 'square root' method... x^2 - 4x + 3 = 0 x^2 - 4x + (4/2)^2 + 3 - (4/2)^2 = 0 (x - 2)^2 + 3 -… [cont.]
Answered by chin_hung148 - Sat Aug 25 23:44:09 2007
Q. I have 2 equations that I need to solve using the square root property. 2x^2-5=93 and (x+4)^2=81 and I need to explain how I got the answers. How would I do this?
Asked by Creole - Sat Aug 25 23:37:30 2007 - - 2 Answers - 0 Comments
A. To solve a equation using the method of 'square root' in a quadratic equation, the equation must be of the form (x + h)^2 = k. If the equation is not of the form (x + h)^2 = k, you would have to apply 'completing the square' method to manipulate a quadratic equation of the form ax^2 + bx +c = 0 to (x + h)^2 = k. 2x^2 - 5 = 93 2x^2 = 98 x^2 = 49 x = 7 or x = -7 (x + 4)^2 = 81 x + 4 = 9 or x + 4 = -9 x = 5 or x = -13 For these two equations, you can use the 'square root' method straightaway without applying 'completing the square'. Here is an example of a quadratic equation which requires you to 'complete the square' before you can use the 'square root' method... x^2 - 4x + 3 = 0 x^2 - 4x + (4/2)^2 + 3 - (4/2)^2 = 0 (x - 2)^2 + 3 -… [cont.]
Answered by chin_hung148 - Sat Aug 25 23:44:09 2007
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tyresmoke.net
cruiser647
hu, 09 Jul 2009 07:51:15 GM
If you actually look at the "diffusers" on the Clios and Imprezas, they're just part of the design on the rear bumper, so they're doing the . square root. of feck all. The idea of a rear diffuser is to create a vacuum under the rear of the ...
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