Consider a+b=ksqrt(ab) where a, b are unknown and k is a constant (in the given problem k=5). Want to find a/b. Note that a+b=ksqrt(ab) is cyclical equation. It is clear that y≠0. Othetwise it will make a/b undefined. From sqrt(ab), ab>=0 --> both a and b must be +ve or -ve. Let r be the value of a/b. r>0 as a and b have the same sign. r=a/b --> a=rb LHS=a+b =(r+1)y RHS=k×sqrt(ab) =k×sqrt(r) Thus r+1=k×sqrt(r) (1) k=(r+1)/sqrt(r) =sqrt(r)+[1/sqrt(r)] (2) This is the equation relating to the ratio of a/b to k. It is important to formulate a problem: • if the two unknowns a and b are fixed to certain value we know the value k, the constant • for a fixed constant we can calculate a/b • as a away to check the result in computating the value of a/b. Solving a+b=5sqrt(ab) can be done by: • dividing the equation by b, yielding r+1=5sqrt(r) as (1) • dividing the equation by sqrt(ab), yielding sqrt(r)+sqrt(1/r)=5 as (2) • squaring: (a+b)²=5²ab Divide by ab and take the sqrt: [sqrt(r)+sqrt(1/r)]=5 as (2). But what the use of squaring and then taking square root prior to division when in fact we can directly to make the division? But it is temping to do the squaring and then to simplify as a²-23ab+b²=0 and then dividing by b²: r²-23ŕ+1=0 a quadratic equation in r I have not checked the results.
a+b=5sqrt(ab) Divide by b: (a/b)+1=5sqrt(a/b) Note that a/b=[sqrt(a/b)]². It means that a/b is positive. The equation can be rewritten as [sqrt(a/b)[²+1=5sqrt(b/a) Putting all terms in one side [sqrt(a/b)]²-5sqrt(a/b)+1=0, a quadratic equation in sqrt(a/b). sqrt(a/b)=½[5±sqrt(21)] Squaring and simplifying: a/b=½[13+5sqrt(21)] as a/b>0
I am puzzled why I get different anwer from that given in the video. At 2:39 you get (a/b)+(b/a)-23=0. It is a cyclical equatiion: interchanging a/b and b/a we'll get the same answer. Letting n=b/a and solve for n (automatically for b/a) 3:11 we get the same value for n (automatically for b/a). Doesn't it mean that a/b=b/a or a=b?
@@nasrullahhusnan2289 No, that's not the correct conclusion. Rather, the solution set must be inversion symmetric; if x is a solution then so must be 1/x. Which holds. You did mess up your simplification in the end a bit. In your answer, sqrt(a/b)=½[5±sqrt(21)] , a,b > 0 simplifies to a/b = ½[23 ± 5sqrt(21)]. Observe that those 2 solutions you get *are* inverses of one another: 1 /([23 - 5sqrt(21)] / 2) = 2(23 + 5sqrt(21) / ([23 - 5sqrt(21)]*[23 + 5sqrt(21)]) = 2(23 + 5sqrt(21) / (23^2 - 25 * 21) = (23 + 5sqrt(21) / 2 . If you solve for n = b/a, you get the same solutions, just in "different order". I should point out that the answer to the original problem also includes a/b being undefined. As a, b = 0 is a valid solution to a+b=5sqrt(ab) .
@@nasrullahhusnan2289 Exactly what i said. The question reads: assume a+b = 5sqrt(ab), what is a/b. The answer to which is either undefined (a and b = 0) or a/b is one of the 2 solutions found.
Consider a+b=ksqrt(ab) where a, b are unknown and k is a constant (in the given problem k=5). Want to find a/b.
Note that a+b=ksqrt(ab) is cyclical equation.
It is clear that y≠0. Othetwise it will make a/b undefined.
From sqrt(ab), ab>=0 --> both a and b must be +ve or -ve.
Let r be the value of a/b. r>0 as a and b have the same sign.
r=a/b --> a=rb
LHS=a+b
=(r+1)y
RHS=k×sqrt(ab)
=k×sqrt(r)
Thus r+1=k×sqrt(r) (1)
k=(r+1)/sqrt(r)
=sqrt(r)+[1/sqrt(r)] (2)
This is the equation relating to the ratio of a/b to k. It is important to formulate a problem:
• if the two unknowns a and b are fixed to certain value we know the value k, the constant
• for a fixed constant we can calculate a/b
• as a away to check the result in computating the value of a/b.
Solving a+b=5sqrt(ab) can be done by:
• dividing the equation by b, yielding r+1=5sqrt(r) as (1)
• dividing the equation by sqrt(ab), yielding sqrt(r)+sqrt(1/r)=5 as (2)
• squaring: (a+b)²=5²ab
Divide by ab and take the sqrt:
[sqrt(r)+sqrt(1/r)]=5 as (2). But what the use of squaring and then taking square root prior to division when in fact we can directly to make the division?
But it is temping to do the squaring and then to simplify as
a²-23ab+b²=0
and then dividing by b²:
r²-23ŕ+1=0
a quadratic equation in r
I have not checked the results.
a+b=5sqrt(ab)
Divide by b: (a/b)+1=5sqrt(a/b)
Note that a/b=[sqrt(a/b)]². It means that a/b is positive. The equation can be rewritten as
[sqrt(a/b)[²+1=5sqrt(b/a)
Putting all terms in one side
[sqrt(a/b)]²-5sqrt(a/b)+1=0, a quadratic equation in sqrt(a/b).
sqrt(a/b)=½[5±sqrt(21)]
Squaring and simplifying:
a/b=½[13+5sqrt(21)] as a/b>0
I am puzzled why I get different anwer from that given in the video.
At 2:39 you get (a/b)+(b/a)-23=0. It is a cyclical equatiion: interchanging a/b and b/a we'll get the same answer. Letting n=b/a and solve for n (automatically for b/a) 3:11 we get the same value for n (automatically for b/a). Doesn't it mean that a/b=b/a or a=b?
@@nasrullahhusnan2289 No, that's not the correct conclusion. Rather, the solution set must be inversion symmetric; if x is a solution then so must be 1/x. Which holds.
You did mess up your simplification in the end a bit. In your answer, sqrt(a/b)=½[5±sqrt(21)] , a,b > 0 simplifies to a/b = ½[23 ± 5sqrt(21)].
Observe that those 2 solutions you get *are* inverses of one another: 1 /([23 - 5sqrt(21)] / 2) = 2(23 + 5sqrt(21) / ([23 - 5sqrt(21)]*[23 + 5sqrt(21)]) = 2(23 + 5sqrt(21) / (23^2 - 25 * 21) = (23 + 5sqrt(21) / 2 . If you solve for n = b/a, you get the same solutions, just in "different order".
I should point out that the answer to the original problem also includes a/b being undefined. As a, b = 0 is a valid solution to a+b=5sqrt(ab) .
@Peikotus91 : It's true that (a,b)=(0,0) satisfy a+b=5sqrt(ab). But recall that the question is a/b. It'll be undefined.
@@nasrullahhusnan2289 Exactly what i said. The question reads: assume a+b = 5sqrt(ab), what is a/b. The answer to which is either undefined (a and b = 0) or a/b is one of the 2 solutions found.
Thanks for the challenge, which I solved 5 steps - with no substitutions. I simply solved for a quadratic in (a/b).
Dividiendo por b a ambos lados y haciendo m = a/b, se obtiene
m + 1 = 5*sqrt(m)
(m + 1)^2 = 25m
m^2 + 2m + 1 = 25m
m^2 - 23m + 1 = 0
m = (23 +- sqrt(23^2 - 4)) /2
= (23 +- sqrt((25-2)^2 - 4))/2
= (23 +- sqrt(25^2-4*25+4-4))/2
= (23 +- 5*sqrt(21))/2
√525 is not exactly equal to 22.9, it is better to simplify by √(25×21)=5√21.
Easy
الرياضيات ليست حسابا فقط. هناك الروابط المنطقية و الشروط.....
Oxford Entrance Math problem: a + b = 5√ab; a/b =?
a, b ≠ 0; (a + b)/b = (5√ab)/b, a/b + 1 = 5√(a/b), [√(a/b)]² - 5√(a/b) + 1 = 0
√(a/b) = (5 ± √21)/2, a/b = [(5 ± √21)/2]² = (46 ± 10√21)/4 = (23 ± 5√21)/2
Answer check:
a + b = 5√ab, a/b + 1 = 5√(a/b)
a/b = (23 ± 5√21)/2; a/b + 1 = (23 ± 5√21)/2 + 1 = (25 ± 5√21)/2
a/b = [(5 ± √21)/2]²; 5√(a/b) = 5[(5 ± √21)/2] = (25 ± 5√21)/2; Confirmed
Final answer:
a/b = (23 + 5√21)/2 = 22.956 or x = (23 - 5√21)/2 = 0.044
The calculation was achieved on a smartphone with a standard calculator app
a+b=5 root a.b = 1+4=5 root 4=2
0,10+4,90=5 root 49 = 5 root 7
I dont know ❤❤
Just k=a/b
a & b are eliminated
k2-23k+1=0
Solve k
No challenge at all😂
(a+b)^2=25ab; i,e a^2+2ab+b^2=25ab ; i,e (a^2/ab)+(2ab/ab)+(b^2/ab)=25 ; i,e )(a/b)+(b/a)+2=25; (b/a)+(a/b)=23 ; Let a/b=x ; So, x+(1/x)=23 ; i,e x^2+1=23x ; i,e x^2-23x+1=0 ; i,e x= [23+√(23^2-4)]/2; or [23-√(23^2-4)]/2 ;; i,e x=(23+√525)/2 or (23-√525)/2 ; i,e x= (23+5√21)/2 or (23-5√21)/2. i,e a/b=(23+5√21)/2 or (23-5√21)/2