Right after I finished my first college degree (double major; 2-year computer science with mathematics), I got a lot of paying work as a math tutor. One day, I went back to my old Junior High School to sign-up to be a math tutor. The lady in the office told me to go see the “head” of the Math department. Turned out the department “head” was my old 8th grade math teacher. Right away, he asked me how much “math” I had taken. I rattled off a list, including Differential Equations and Linear Algebra ( *note:* the year was 1986.) My former 8th grade math teacher replied… “WHOA! _they didn’t_ _even require me to_ _take as much as_ LINEAR ALGEBRA _to_ _get my Masters with_ _teaching certificate!”_ ( *note:* my former 8th grade math teacher probably started teaching the 8th grade around the year 1974.) My discussion with my former 8th grade math teacher took place in 1986. The discussion revealed that I, with only a 2-year degree, took MORE MATH (circa 1983-1986) than my former 8th grade math teacher (who held a Masters and a Teaching Certificate.) In the decades since, the situation has only become “more amplified.” Now, here in 2025, to take and pass an AP (advanced placement) exam, you need to know as much if not more math than I had to know upon completing a 2-year Computer Science with Math degree back in 1986. This outcome is due to how the internet, between 2000 to 2010, made higher math available to any 5th grader with an internet connection. And to think that over 98% of the Higher Math was known to about 0.01% of the Human Race as of the year 1890. Basically, it took time for higher math to “trickle down” to graduate students, and even longer to “trickle down” to 4-year degree students, and then “trickle down” to 2-year degree students, and then, by the year 2008, to finally “trickle down” to 5th graders, thanks to the advent of TH-cam, Wikipedia, and the internet. Wheels within wheels!
Let's do it the multiple-choice, speed-over-accuracy way, just for funsies. 4.something + 6.something = something in the range 10 to 12 a = 7.something b < 1 c = 2.something × 5 so it's in the ballpark d = 1.4... × 7 = 9.8 It's c.
I'm not sure that estimating values for all four of the answer choices and also the expression itself is quicker than doing the actual simplification in this case.
But with multiple choice, its often quicker just to rule out the wrong answers. If you can do this by estimating, you will have more time to attend to other questions. √20 is about 4.5 √45 is about 7 So our answer is about 11 Looking at our possible answers. A: √50 is about 7 (so that's too small) B: Is less than one (so also too small) C: is approximately 11 (so that's promissing) D: is about 10 (probably a bit less so once again too small) So C is my answer within about 5 seconds.
That's definitely a good strategy to remember for multiple choice questions, but do you really think you can calculate and compare five estimations like that quicker than just simplifying the expression in this case?
@@gavindeane3670 I did. Straight from the thumbnail. Like I say, you don't have to calculate anything, you just need an approximation. If it wasn't so cut and dried, then I'd have used other methods. Maybe even the one presented in this video, but I think the approximations are far quicker than the simplification shown here. A) 50 is very close to 49 so at a glance, it's about 7 (no calculation required) B) You know that the square root of 2 is smaller than 2 so 3x√2 is obviously less than 10 so (3x√2) /10 is obviously less than 1. C) √5 is very roughly 2.25 so 5x√5 is obviously around about 11. D) √2 is less than 1.5 so 7x√2 is less than 10.5 You only need to look at each answer for a couple of seconds to come to these conclusions. I think that's quicker than simplifying the expression. It should only take about 5 seconds and 10 at the very most.
@@KenFullmanIt took me less time than that to simplify this to 5√5. Either strategy is fine. I'm just a bit surprised that evaluating five different square root expressions might be faster than adding 2+3.
Alright, so root math. Let’s get to it. Sqrt(20) + sqrt(45) Notice a factor of 5 is common to both of these. So let’s factor down: = sqrt(5 * 4) + sqrt(5 * 9) And because everything is positive, we can do this: = sqrt(4) * sqrt(5) + sqrt(9) * sqrt(5) = 2*sqrt(5) + 3*sqrt(5) I hope I don’t have to explain why 2 and 3 are only positive. Let’s do a last bit of factorizing: = (2+3)sqrt(5) = 5 * sqrt(5) So the answer is C! You could also guess its C by estimation: 1 < sqrt(2) < 2 < sqrt(5) < 3 4 < sqrt(20) < 5 6 < sqrt(45) < 7 sqrt(20) + sqrt(45) ≈ 4.5 + 6.5 = 11 And our possibilities: -a.) sqrt(50) = sqrt(25*2) = 5*sqrt(2) ≈ 5 * 1.5 = 7.5 TOO SMALL -b.) You should be able to just look at this and tell it’s less than a. TOO SMALL -c.) 5*sqrt(5) ≈ 5 * 2.5 = 12.5 PLAUSIBLE -d.) 7*sqrt(2) ≈ 7 * 1.5 = 10.5 PLAUSIBLE This would get you down to C and D as 7*sqrt(2) is deviously close to 11 itself, but you should notice that since 2 is not a common factor to 20 and 45, it’s not likely to be part of the final answer. And before you shout “That’s not precise!”, remember that estimation IS a math skill. Any questions?
Actually, it’s not that simple, especially if you’re not allowed to use a calculator on the test. In order to solve this, you need to either have good estimation skills and good mathematical intuition, or you need to know what you can AND cannot do with square roots. It may look simple, but I guarantee if you showed this to most people, they’d do something wrong.
√20 + √45 ≠ √65
simplify each radical term first before adding
The square numbers :
0² = 0 , 1² = 1, 2² = 4, 3² = 9, 4² = 16, 5² = 25
6² = 36, 7² = 49, 8² = 64, 9² = 81 ...n²
S = {0, 1, 4, 9, 16, 25, 36, 49, 64, 81, ...}
check if 20 and 45 is divisible by the numbers of the set
20 : 4 = 5 --> 20 --> 2² · 5
√20 = √[2² · 5 ] = 2√5
45 : 9 = 5 --> 45 --> 3² · 5
√45 = √[3² · 5] = 3√5
now its possible to add up the radicals
√20 + √45 = 2√5 + 3√5 = 5√5✅ --> (c
2√5 + 3√5 = √5(2 + 3) = 5√5✅
Right after I finished my first college degree (double major; 2-year computer science with mathematics), I got a lot of paying work as a math tutor. One day, I went back to my old Junior High School to sign-up to be a math tutor. The lady in the office told me to go see the “head” of the Math department. Turned out the department “head” was my old 8th grade math teacher. Right away, he asked me how much “math” I had taken. I rattled off a list, including Differential Equations and Linear Algebra ( *note:* the year was 1986.) My former 8th grade math teacher replied…
“WHOA! _they didn’t_ _even require me to_ _take as much as_ LINEAR ALGEBRA _to_ _get my Masters with_ _teaching certificate!”_
( *note:* my former 8th grade math teacher probably started teaching the 8th grade around the year 1974.) My discussion with my former 8th grade math teacher took place in 1986. The discussion revealed that I, with only a 2-year degree, took MORE MATH (circa 1983-1986) than my former 8th grade math teacher (who held a Masters and a Teaching Certificate.) In the decades since, the situation has only become “more amplified.” Now, here in 2025, to take and pass an AP (advanced placement) exam, you need to know as much if not more math than I had to know upon completing a 2-year Computer Science with Math degree back in 1986. This outcome is due to how the internet, between 2000 to 2010, made higher math available to any 5th grader with an internet connection. And to think that over 98% of the Higher Math was known to about 0.01% of the Human Race as of the year 1890. Basically, it took time for higher math to “trickle down” to graduate students, and even longer to “trickle down” to 4-year degree students, and then “trickle down” to 2-year degree students, and then, by the year 2008, to finally “trickle down” to 5th graders, thanks to the advent of TH-cam, Wikipedia, and the internet. Wheels within wheels!
Thank you
(2✓5) + (3✓5)=5✓5
√20 + √45 = 2√5 + 3√5 = 5√5 = c.
sqrt(20)
+
sqrt(45)
=
sqrt(20)=sqrt(4×5) =2sqrt(5)
sqrt(45)=sqrt(9×5) =3sqrt(5)
added togerher =5sqrt(5)
V20 = V(2 . 2 . 5) = 2V5 and V45 = V(3 . 3 . 5) = 3V5 so V20 + V45 = 2V5 + 3V5 = 5V5
Let's do it the multiple-choice, speed-over-accuracy way, just for funsies.
4.something + 6.something = something in the range 10 to 12
a = 7.something
b < 1
c = 2.something × 5 so it's in the ballpark
d = 1.4... × 7 = 9.8
It's c.
I'm not sure that estimating values for all four of the answer choices and also the expression itself is quicker than doing the actual simplification in this case.
@@gavindeane3670 Nor am I, but it's definitely a close race. And either way it takes a lot longer to type than to do.
c) 5 sq.root of 5
got 5 SR 5 each reduces by its sq factor. 2 SR 5 + 3 SR 5 = 5 SR 5 thanks for the fun.
But with multiple choice, its often quicker just to rule out the wrong answers. If you can do this by estimating, you will have more time to attend to other questions.
√20 is about 4.5
√45 is about 7
So our answer is about 11
Looking at our possible answers.
A: √50 is about 7 (so that's too small)
B: Is less than one (so also too small)
C: is approximately 11 (so that's promissing)
D: is about 10 (probably a bit less so once again too small)
So C is my answer within about 5 seconds.
That's definitely a good strategy to remember for multiple choice questions, but do you really think you can calculate and compare five estimations like that quicker than just simplifying the expression in this case?
@@gavindeane3670 I did. Straight from the thumbnail. Like I say, you don't have to calculate anything, you just need an approximation. If it wasn't so cut and dried, then I'd have used other methods. Maybe even the one presented in this video, but I think the approximations are far quicker than the simplification shown here.
A) 50 is very close to 49 so at a glance, it's about 7 (no calculation required)
B) You know that the square root of 2 is smaller than 2 so 3x√2 is obviously less than 10 so (3x√2) /10 is obviously less than 1.
C) √5 is very roughly 2.25 so 5x√5 is obviously around about 11.
D) √2 is less than 1.5 so 7x√2 is less than 10.5
You only need to look at each answer for a couple of seconds to come to these conclusions. I think that's quicker than simplifying the expression. It should only take about 5 seconds and 10 at the very most.
@@KenFullmanIt took me less time than that to simplify this to 5√5.
Either strategy is fine. I'm just a bit surprised that evaluating five different square root expressions might be faster than adding 2+3.
Alright, so root math. Let’s get to it.
Sqrt(20) + sqrt(45)
Notice a factor of 5 is common to both of these. So let’s factor down:
= sqrt(5 * 4) + sqrt(5 * 9)
And because everything is positive, we can do this:
= sqrt(4) * sqrt(5) + sqrt(9) * sqrt(5)
= 2*sqrt(5) + 3*sqrt(5)
I hope I don’t have to explain why 2 and 3 are only positive. Let’s do a last bit of factorizing:
= (2+3)sqrt(5)
= 5 * sqrt(5)
So the answer is C! You could also guess its C by estimation:
1 < sqrt(2) < 2 < sqrt(5) < 3
4 < sqrt(20) < 5
6 < sqrt(45) < 7
sqrt(20) + sqrt(45) ≈ 4.5 + 6.5 = 11
And our possibilities:
-a.) sqrt(50) = sqrt(25*2) = 5*sqrt(2) ≈ 5 * 1.5 = 7.5 TOO SMALL
-b.) You should be able to just look at this and tell it’s less than a. TOO SMALL
-c.) 5*sqrt(5) ≈ 5 * 2.5 = 12.5 PLAUSIBLE
-d.) 7*sqrt(2) ≈ 7 * 1.5 = 10.5 PLAUSIBLE
This would get you down to C and D as 7*sqrt(2) is deviously close to 11 itself, but you should notice that since 2 is not a common factor to 20 and 45, it’s not likely to be part of the final answer. And before you shout “That’s not precise!”, remember that estimation IS a math skill.
Any questions?
Awesome thanks 👍🙏👏💪😎🌎
Did you know this?: (20 + 25)^2=2025
Cool math in 2025?
Answer: 5 √5
-----------
√20 + √45
√5 x √4. + √5 x √9
2 √5 + 3 √5
5 √5
(e) I had no idea. Thanks for explaining. And nobody worry, I’m not going to be a teacher.
SImple at a glance.
At least it is not another of those ludicrous PEMDAS messes.
C
Such an extremely simple question in a test to become a math teacher. Sad.
Actually, it’s not that simple, especially if you’re not allowed to use a calculator on the test. In order to solve this, you need to either have good estimation skills and good mathematical intuition, or you need to know what you can AND cannot do with square roots. It may look simple, but I guarantee if you showed this to most people, they’d do something wrong.
John the JACKASS, perfect ability to make a mountain of a molehill
C
C