NewStats: 3,263,762 , 8,181,321 topics. Date: Saturday, 07 June 2025 at 07:31 PM 1x2e1

Surds are very common with respect to mathematical basics and very important as not all roots of numbers can be expressed as entirely whole numbers hence, the reason for the concept of "surds".


Ladies and gentlemen, I present to you a simple but trickish question on surds below. The first person to solve it correctly gets a prize. Goodluck!

Cc: olawalebabs, Richiez, Fynestboi.

calculator please

PapiDShipCapt:
calculator please

Answer must be in surd form and you must show workings as well to get the prize, so, calculator won't be useful.

10

alabiyemmy:
10

No. Try again.

Please find the solution attached below.
The answer is approximately 7.1 not 10.

Kindly ignore the 'or 10' in my answer.

Happy New Year

Wey the price Ooo?

Watching

TRADELYN:
Please find the solution attached below.
The answer is approximately 7.1 not 10.

Kindly ignore the 'or 10' in my answer.

Happy New Year

Wey the price Ooo?

Happy new year to you too. You tried but your approach is wrong. You see, squaroot(a+b) is not the same thing as squaroot(a) + squaroot(b). E.g squaroot (15+10) = squaroot(25)= 5 but squareroot (15) + squaroot (10)= 7.035 which is not the same as 5. Try again.

naturalwaves:

Happy new year to you too. You tried but your approach is wrong. You see, squaroot(a+b) is not the same thing as squaroot(a) + squaroot(b). E.g squaroot (15+10) = squaroot(25)= 5 but squareroot (15) + squaroot (10)= 7.035 which is not the same as 5. Try again.

If you say so, then isn't it correct if I take just one square root for all before simplying the figures? If true, I will get same answer.

No space here to re~do the workings now.

^^^ That's not the answer. Show it to your friends.

for how much ?

I have the solution here .

agentofchange1:
for how much ?

I have the solution here .

Hahaha benbucks, I knowingly didn't mention the maths gurus in the house. The prize is just a recharge card token if it is correct.

naturalwaves:

Hahaha benbucks, I knowingly didn't mention the maths gurus in the house. The prize is just a recharge card token if it is correct.

OK

its. 2√15



solution loading

agentofchange1:


OK

its. 2√15



solution loading

Hmmm.

naturalwaves:

Hmmm.

Solution

let's pick √(25+10√6)

put √p +√q =√(25+10√6)

taking square of both sides , we have

p+q +2√(pq) =25 +10√6

compare both sides to have

p+q =25 ....... B

& √(pq)=5√6......E

take square again from E

=>pq=150.....N

from N, p=150/q

put in B

=>q² -25q +150=0

solve to obtain q = 10 ,15

=> p =15..(when q=10)
& p=10....(when q=15)

thus , √(25+10√6)= √15+√10

similarly if we pick √(25-10√6 ) we have √15-√10

hence √(25+10√6)+√(25-10√6)=√15+√10+√15-√10=2√15

that's. it man


guess am right

agentofchange1:


Solution

let's pick √(25+10√6)

put √p +√q =√(25+10√6)

taking square of both sides , we have

p+q +2√(pq) =25 +10√6

compare both sides to have

p+q =25 ....... B

& √(pq)=5√6......E

take square again from E

=>pq=150.....N

from N, p=150/q

put in B

=>q² -25q +150=0

solve to obtain q = 10 ,15

=> p =15..(when q=10)
& p=10....(when q=15)

thus , √(25+10√6)= √15+√10

similarly if we pick √(25-10√6 ) we have √15-√10

hence √(25+10√6)+√(25-10√6)=√15+√10+√15-√10=2√15

that's. it man


guess am right

Infact, over correct dey worry your solution. Thumbs up man. That's the answer. Send a PM to get your gift. Your approach was even totally different from mine but we got the same answer. That's the beauty of mathemtics. See mine below;

naturalwaves:


Infact, over correct dey worry your solution. Thumbs up man. That's the answer. Send a PM to get your gift. Your approach was even totally different from mine but we got the same answer. That's the beauty of mathemtics. See mine below;

hmmm.OK. sir .

^^^ copied. You can remove your number now. Agentofchange, have you checked my matrix solution?

supposed:
√(25+10√6)+√(25–10√6)=x√y;
square everything you will get;
(x√y)²= [√(25+10√6)+√(25–10√6)]²;
(x√y)²= [√(25+10√6)+√(25-10√6)] ×[√(25+10√6)+√(25–10√6)]
(x√y)²= √[(25+10√6)(25+10√6)] +2√[(25+10√6)(25–10√6)] +√[(25–10√6)(25–10√6)];
(x√y)²= 25+10√6 +2√[(25+10√6)(25–10√6)] +25-10√6........ (by √(a×a)=a
(x√y)²= 25+25+10√6–10√6 +2√[25² +25(10√6)–25(10√6) –(10√6)²];
(x√y)²= 50+2√(625–100×6);
(x√y)²=50+2√25=50+2×5=60;
Square root both side of the equation to get x√y;
x√y= √60=√(4×15);
finally;
x√y= 2√15
So therefore;
√(25+10√6)+√(25–10√6)= 2√15.............. In simpler form, I came so late

icecalm:
[size=20]supposed:
√(25+10√6)+√(25–10√6)=x√y;
square everything you will get;
(x√y)²= [√(25+10√6)+√(25–10√6)]²;
(x√y)²= [√(25+10√6)+√(25-10√6)]×[√(25+10√6)+√(25–10√6)]
(x√y)²= √[(25+10√6)(25+10√6)]+2√[(25+10√6)(25–10√6)]+√[(25–10√6)(25–10√6)];
(x√y)²= 25+10√6+2√[(25+10√6)(25–10√6)]+25-10√6........ (by √(a×a)=a
(x√y)²= 25+25+10√6–10√6+2√[25²+25(10√6)–25(10√6)–(10√6)²];
(x√y)²= 50+2√(625–100×6);
(x√y)²=50+2√25=50+2×5=60;
Square root both side of the equation to get x√y;
x√y= √60=√(4×15);
finally;
x√y= 2√15
So therefore;
√(25+10√6)+√(25–10√6)= 2√15.............. In simpler form, I came so late[/size]

Welldone bro. You really tried. Keep it up. Follow this thread, I will paste another one later.

naturalwaves:


Welldone bro. You really tried. Keep it up. Follow this thread, I will paste another one later.

okay

naturalwaves:
Surds are very common with respect to mathematical basics and very important as not all roots of numbers can be expressed as entirely whole numbers hence, the reason for the concept of "surds".


Ladies and gentlemen, I present to you a simple but trickish question on surds below. The first person to solve it correctly gets a prize. Goodluck!

Cc: olawalebabs, Richiez, Fynestboi.

I will solve it,but you must keep to your words and the recharge card as you said.

No need the answer is already there...

naturalwaves:
Surds are very common with respect to mathematical basics and very important as not all roots of numbers can be expressed as entirely whole numbers hence, the reason for the concept of "surds".


Ladies and gentlemen, I present to you a simple but trickish question on surds below. The first person to solve it correctly gets a prize. Goodluck!

Cc: olawalebabs, Richiez, Fynestboi.