1 Olympiad Math Q. For Everyone

[rahulpathak]

a(2b+1) + 0.5( 2b+1) = 8+0.5
(2b+1)(a+0.5) =17/2
(b+0.5)(a+0.5)=17/4
Thus it's rectangular Hyperbola in a-b plane with asymptotes @ a= -0.5, b= -0.5 and C = √17 / 2
Thus equation of directrix
a+b = √(17/2) -1
And a+b = -√(17/2) -1 😊.
NB: Solution of question is in reply section of comment .

[arnavkumar]

Here's how to solve the problem:
* Rearrange the equation:
a + b + 2ab = 8
* Add 1 to both sides:
a + b + 2ab + 1 = 9
* Factor the left side:
Notice that the left side can be factored. It looks very similar to the expansion of (1+a)(1+b):
(1 + a)(1 + b) = 1 + a + b + ab
However we have 2ab instead of ab, but we can still factor by grouping.
a + b + 2ab +1 = (a+1) + b(2a+1)
This doesn't seem to work, so let's try something else.
Let's try to rewrite the equation in a way that allows us to factor.
a + 2ab + b + 1 = 9
a(1+2b) + (b+1) = 9
This still doesn't factor easily.
Let's try to manipulate the equation to look like (1+2a)(1+2b)
(1+2a)(1+2b) = 1 + 2a + 2b + 4ab
We have a+b+2ab=8.
Multiply the equation by 2:
2a+2b+4ab=16
Add 1 to both sides:
2a+2b+4ab+1=17
(1+2a)(1+2b)=17
* Rewrite a+b+2ab+1:
a+b+2ab+1
Rearrange to get:
2ab+a+b+1
Multiply by 2
4ab+2a+2b+2 = 18
4ab+2a+2b+1=17
(2a+1)(2b+1)=17
* Let x = a+b:
We have a+b+2ab=8.
We have (2a+1)(2b+1)=17
4ab+2a+2b+1=17
4ab+2(a+b)+1=17
4ab+2x+1=17
4ab=16-2x
2ab=8-x
* Substitute 2ab into the original equation:
a+b+2ab=8
x + (8-x) = 8
8=8
* Solve for a+b:
a+b+2ab=8
2ab=8-(a+b)
(2a+1)(2b+1)=17
4ab+2a+2b+1=17
4ab+2(a+b)=16
4ab=16-2(a+b)
2ab=8-(a+b)
Let a+b=x
2ab=8-x
4ab=2(8-x)=16-2x
16-2x+2x=16
(2a+1)(2b+1)=17
4ab+2(a+b)+1=17
4ab+2x+1=17
4ab=16-2x
2ab=8-x
a+b+2ab=8
x+8-x=8
8=8
* Alternative approach:
(2a+1)(2b+1) = 17
4ab + 2a + 2b + 1 = 17
4ab + 2(a+b) = 16
2ab + (a+b) = 8
2ab = 8 - (a+b)
Let x = a+b
2ab = 8 - x
4ab = 16 - 2x
16 - 2x + 2x = 16
Therefore, we cannot determine a+b.
However, we can determine that a+b+2ab=8 and (2a+1)(2b+1)=17.
Final Answer:
The value of a+b cannot be uniquely determined from the given equation.

[sohitgamingyt]

a+b+2ab=8
a(1+2b)+b=8
Now we can find values of a+b by keeping right side value as prime number and factorising
a(1+2b)+b+1/2=8+1/2
a(1+2b) + (b+1/2) = 17/2
Multiplying by 2 on Both sides
2a(1+2b) + 1(2b+1) = 17

(2a+1)(2b+1)=17
So case one:2a+1=17, 2b+1=1
So values of a and b are 8 and 0 respectively
Case two:2a+1=1, 2b+1=17
So values of a and b are 0 and 8 respectively

So in both cases value of a + b = 8.

There is one small mistake in the question. It should be given that:
a and b are positive integers.

[kalaneupane]

Let the given equation be
a+b+2ab = 8
We want to find a+b. Let x = a+b. Then the equation becomes
x + 2ab = 8
We need to express ab in terms of x.
We can rewrite the given equation as
a+b+2ab+1 = 8+1
a+b+2ab+1 = 9
a(1+2b) + (b+1) = 9
We can also write
2ab + a + b = 8
4ab + 2a + 2b = 16
4ab + 2a + 2b + 1 = 17
(2a+1)(2b+1) = 17
Let 2a+1 = u and 2b+1 = v. Then uv = 17.
We have 2a = u-1 and 2b = v-1.
a = \frac{u-1}{2} and b = \frac{v-1}{2}.
Then a+b = \frac{u-1}{2} + \frac{v-1}{2} = \frac{u+v-2}{2}.
Also, ab = \frac{(u-1)(v-1)}{4} = \frac{uv - u - v + 1}{4}.
Since uv=17, we have ab = \frac{17 - u - v + 1}{4} = \frac{18 - (u+v)}{4}.
We have a+b+2ab = 8.
So, \frac{u+v-2}{2} + = 8.
\frac{u+v-2}{2} + \frac{18-(u+v)}{2} = 8
\frac{u+v-2+18-u-v}{2} = 8
\frac{16}{2} = 8
8 = 8
This equation is always true when uv=17.
However, we want to find a+b.
Let a+b = x.
Then x + 2ab = 8.
2ab = 8-x.
ab = \frac{8-x}{2}.
(2a+1)(2b+1) = 17
4ab + 2a + 2b + 1 = 17
4ab + 2(a+b) + 1 = 17
4\left(\frac{8-x}{2}\right) + 2x + 1 = 17
2(8-x) + 2x + 1 = 17
16-2x+2x+1 = 17
17 = 17
This equation is true for all x.
However, we want to find a unique value for a+b.
Let's assume a and b are integers. Since 17 is prime, we have the following possibilities for (2a+1, 2b+1):
(1, 17), (17, 1), (-1, -17), (-17, -1).
Case 1: 2a+1 = 1, 2b+1 = 17. Then a=0, b=8. a+b = 8.
Case 2: 2a+1 = 17, 2b+1 = 1. Then a=8, b=0. a+b = 8.
Case 3: 2a+1 = -1, 2b+1 = -17. Then a=-1, b=-9. a+b = -10.
Case 4: 2a+1 = -17, 2b+1 = -1. Then a=-9, b=-1. a+b = -10.
If a+b=8, 8+2ab=8, so 2ab=0. Thus a=0 or b=0.
If a+b=-10, -10+2ab=8, so 2ab=18, ab=9.
If a=-1, b=-9, then ab=9.
If a=-9, b=-1, then ab=9.
Thus, a+b can be 8 or -10.
Final Answer: The final answer is \boxed{8}

[navmeena]

Le SSC student : 1 equation 2 variable hai value putting laga do… put B = 0 which gives A = 8 then A + B = 8.

[jaydeepkumar]

We are given the equation:
a + b + 2ab = 8
and we need to find, given that and are positive numbers.

Step 1: Let

Rewriting the given equation:

S + 2ab = 8

Using the identity:

(a + b)^2 = a^2 + b^2 + 2ab

we express as:

a^2 + b^2 = S^2 - 2ab

Since we already know that, solving for :

2ab = 8 - S

Substituting in the identity:

S^2 - 2ab = S^2 - (8 - S)

which simplifies to:

S^2 - 8 + S = 0

Step 2: Solve for

Rearranging:

S^2 + S - 8 = 0

Solving this quadratic equation using the quadratic formula:

S = (-1± √(1 + 32))/2

Approximating :

S = (-1± 5.744)/(2)

S = 4.744/2

S = 2.37 and -3.37

Since and are positive, we take the positive value:

a + b ≈ 2.37

So, the final answer is

a+b =( -1 + √33)/2

[sumanchakraborty]

We need conditions defined about the property of a and b to solve and obtain particular values of a and b.
Without the properties properly defined, there can be multiple values for both a and b

[Lol-ssyf]

Infinite solutions
a+b+2ab=8
a+2ab=8-b
a(1+2b)=8-b
a=(8-b)/(1+2b)
Now for any value of b (except -1/2) there will exist a unique value of a. Thus a+b does not have a definite answer.

For example, for B=0, A=8
for B=1, A=7/3
for B=2, A=6/5, etc.

[MrDc]

a + b + 2ab = 8
a(1 + 2b) + b = 8
a(1 + 2b) + b + 1/2 = 8 + 1/2
a(1 + 2b) + (b + 1/2) = 17/2
2a(2b + 1) + (2b + 1) = 17
(2b + 1)(2a + 1) = 17 [Note :- 17 = (- 17) × (- 1)]

If (2a + 1) > (2b + 1) = 17 > 1 or (2a + 1) < (2b + 1) = 1 < 17
2a + 1 = 17 or 2a + 1 = 1
a = 8 😀 or a = 0 😀

2b + 1 = 1 or 2b + 1 = 17
b = 0 😀 b = 8 😀

Therefore a+ b = 8 ✔✔✔

If (2a + 1) > (2b + 1) = -1 > -17 or (2a + 1) < (2b + 1) = -17 < -1
2a + 1 = - 1 or 2a + 1 = - 17
a = - 1 😀 or a = - 9 😀

2b + 1 = - 17 or 2b + 1 = - 1
b = - 9 😀 b = - 1 😀

Therefore a+ b = - 10 ✔✔✔

[naveensharma]

a+b+2ab = 8
(a+b)+2ab = x+(8-x)
2ab = 8-x
2ab is always even. Therefore 8-x must be even.
Possible solutions:
Case I: x = 0
2ab = 8
a+b = 0
=> a= +or-2, b= -or+2

Case II: x=2
2ab = 8-2 = 6
a+b = 2
a= imaginary root by quadratic equation, b same

Case III: x=4
2ab = 8-4 = 4
a+b = 4
a= imaginary root of the quadratic equation, b same

Case IV: x = 6
2ab = 8-6= 2
a+b = 6
a= 3/2+(or-)√5/2, b= 9/2-(or+)√5/2

Case V: x=8
2ab= 8-8=0
a+b= 8
a= 8 or 0, b = 0 or 8.

You can go for -ve also but if only a positive integer solution is required, then only one solution fits that is case V where a and b are 0 and 8 or vice versa.
Otherwise many solutions are there for this case depending on the options of MCQ.

[ManavWayal]

Here is correct solution
a+2ab+b=8,
a(1+2b)+b=8,
a(1+2b)+1/2(1+2b)=8+1/2,
2a(1+2b)+(1+2b)=16+1,
(2a+1)(1+2b)=17,
17 is a prime number, there fore we can make 17 in only two ways 1×17 and 17×1,
1.
2a+1=1. 1+2b=17
(a, b)=(0, 8)
2.
2a+1=17. 1+2b=1
(a, b)=(8, 0)

[farzanakhandokar]

Due to the time limit in Olympiad... The fastest approach to this answer is either : 1) Dividing the given value by (in this question it's 8) into 2 equal portions 0r maybe uneven portions & then, try to set those value manipulatively into 'a' & 'b' variables.

Or

2) Set one variable's value to 0
If, we calculate in the 1st method: then, (a + b + 2ab) = 8 2 + 2 + 2(2 x 2) = 12....So, in this case, our answer is wrong!

However, even if we take 'a' =1 & 'b' = 3 ..still, our answer will be wrong(it will be 10)

...So, we'll take an attempt in the 2nd method... Let's try it! let's set 'a' = 8 & 'b' = ( a+ b + 2ab ) = 8...therefore, 8 + 0 +2(8 x 0) = 8..
As we're seeing in this case (a + b) = 8...Doesn't matter if we're setting 'a' = 0 or 'b' = 0....

Hence, you can say that it is our only limitation in this case, as we can't determine the exact value of 'a' or 'b'! However, as I mentioned: "The Fastest Approach" ...
Therefore, You can say it a trick to solve faster!... If you think I did something wrong please let me know! & If you know a better solution also tell me by replying to this comment!

[DEEPSUM-id]

a+b+2ab = 8

To find a + b, we can approach it by trying to factor the equation.

Let's denote x = a + b and y = ab.

We can rewrite the given equation as:

x + 2y 8

Now, x a+b and y = ab. This equation alone is not enough to determine the exact values of a and b, but we can try to find relationships between x and y.

If we treat this as a quadratic equation, the two numbers a and b satisfy the equation:

t² (a+b)t + ab = 0

This means that the quadratic equation for a and b is:

t² - xt + y = 0

For real solutions to exist for a and b, the discriminant of this quadratic equation must be non-negative. The discriminant is given by:

A = x² - 4y 2

Substitute y = 2 from the equation x + 2y 8 into the discriminant condition:

A = x²-4 2 8-x 2

Simplifying:

A = x²-2(8x) = x²-16+2x = x² + 2x - 16

For real solutions, the discriminant must be non-negative:

x²+2x-16≥0

We solve this quadratic inequality:
For real solutions, the discriminant must be non-negative:

x²+2x-16 ≥ 0

We solve this quadratic inequality:

x² + 2x - 16 0

Using the quadratic formula:

-222-4(1)(-16) 2(1) X -2+4+64 -2±√68 = -2±2/17 2

x=-1+√17

Thus, the possible values for x = a + b are:

X -1+√17 or x = -1 √17

These are the possible values for a + b, so:

a+b=-1+ √17 or a+b=-1-17

These are the two potential values of a + b.

[shorovybristy]

a+b+2ab=8
a+b=?
Solution :
(a+b)^2 =8^2
or, √ (a+b)^2 =√ 64
‌therefore, a+b=8


In this process I did it

[tbapee]

Actually in these types of MCQ questions, it is better to hit and trial the answer than to calculate it if possible due to less time availability. Here a=8 and b=0 supports the equations. So a+b=8

[subhajitruidas]

a+b+2ab=8
Or, (√a)^2+ (√b)^2 + 2*( √a)*(√b)=8
Or, (√a+√b)^2=8
Or, √a+√b= +-(2√2)
Now root always give positive value
Then
√a+√b must be Positive
Hence
√a+√b=2√2
the equan hold if you put a=2 and b=2
Therefore a+b=4
( Also another ans possible)
But 4 is best ans the equation is in symmetry so the value of a and b is equal

[RupeshMahanta_]

a+b+2ab=8
a+b=

Squaring both side
(a+b)²=(8-2ab)²
(a+b)²= 8²-2.8.2ab+ (2ab)²
(a+b)²= 64-32ab+4(ab)²
(a+b)²= 4(ab)²-32ab+64
(a+b)²= (ab)²-8ab+16 (Quadratic)
(a+b)²= (ab-4)(ab-4)

ab= 4, now substituting in equation (1)

a+b= 8-2ab
a+b= 8-2×4
a+b= 0

[Alphamatics]

a+b+2ab =8
common trick
a(2b+1 ) +1/2(2b+1)-1/2 =8
(a+1/2)(2b+1) = 17/2
(2a+1)(2b+1) =17
17=17×1or -17×-1
a=8
b =0
Or, a= 0 b=8
And a= -9
b = -1 or vise - versa
The - ve case don't satisfy the above equ
So a+b = 8

[KartikBargali-ye]

दिया है कि a+ b+2ab=8

हल - a+b+2ab=8 is eqn -1
a+2ab+b=8
a(1+2b)+b=8
a(1+2b)=8-b
a=8-b/(1+2b) is eqn 2
Put value of a in eq1 we get

8-b/1+2b+b+2(8-b/1+2b).b=8
8-b/1+2b+2b(8-b/1+2b)+b=8

{8-b+2b(8-b)/1+2b}+b=8



4b square+16b+8=8(1+2b)
4b square+16b+8=8+16b

4b square=0
So b=0
Put b vale in eqn 2 we get
a=8-(0)/1+2(0)
a=8

So( a=8), (b=0)
so a+ b=8

Mene bahut kosis ki hai explain karne mai in comments, ager galat hoga to maf kar dena par i did try 😂😅

[kashifahmead]

if, a + b +2ab = 8
=> (a+b)^2 = 8
=> rooting both sides
=> (a+b) = 2root(2) [calculate krlena 8 ka root]
Hence the the value of a+b is 2root(2) 😎✅