The second solution for the equation 2^a+4^b+8^c=584 is a=9, b=3, c=1. 584 have also 512, 64, and 8 as component of the sum. 512 and 8 can be expressed as 2^9 resp. 8^1 or 8^3 resp.2^3. 64 remains in both solutions 4^3.
Your Video was very helpful and very nice for education. But I have a problem in some questions and I am trying hard to hard but not able to find exact answers of variables. Can you solve those for me ? Q. 1 Given - a - b = 2 , a^5 - b^5 = 992 a = ? , b = ? Q. 2. Given - 25x^2 + 49y^2 = 841 , xy = 12 , x = ? , y = ? Q. 3. Given - (x^2 - 3x)^2 - 8(x^2 - 3x) = 20 , x = ? Q. 4. Given - x^4 - x^2 = -1 , x = ? Please , Solve these questions and help me . Solve when ( or if ) you are free .
Q. 1 Given - a - b = 2 , a^5 - b^5 = 992 a = ? , b = ? From the symmetry of the given equations it follows that if (a,b)=(x,y) is a solution, then (a,b)=(-y,-x) is also a solution. From the first equation, b=a-2, so substitute a-2 for b in the first equation: a^5-(a-2)^5=992. Expand and simplify to obtain a^4-4a^3+8a^2-8a-96=0. f(a)=a^4-4a^3+8a^2-8a-96 is a convex function (the 2nd derivative is >0 for all a), so there can be at most two real solution for a. a=-2 is such a solution, and from the first equation we have b=-4. From the symmetry of the equation we have that a=4, b=2 is the second real solution. If you want the complex solutions, simply divide f(a) by (a+2)(a-4) to obtain a quadratic equation for the remaining complex solutions and solve using the well known formula for quadratic equations. Q. 2. Given - 25x^2 + 49y^2 = 841 , xy = 12 , x = ? , y = ? y^2=144/x^2 from the second equation. Use this in the first equation and multiply by x^2 to obtain a quadratic equation for z=x^2: 25z^2-841z+49*144. Use the well known formula to obtain z=[841±√(841^2-4*49*144*25)]/50. This simplifies by elementary arithmetic to z=16 or z=441/50. Therefore x=±√z, or x=±4 or x=±21/5. Use the second equation to find the corresponding values for y. Q. 3. Given - (x^2 - 3x)^2 - 8(x^2 - 3x) = 20 , x = ? Let y=(x^2 - 3x). The given equation becomes a simple quadratic equation in y. Use the well known formula bla bla bla to obtain y=-2 or y=10. Then solve the two quadratic equations x^2 - 3x+2=0 and x^2 - 3x+2-10=0 to find the for solutions [1, 2, -2, 5]. Q. 4. Given - x^4 - x^2 = -1 , x = ? This has 4 complex and no real solutions. Let y=x^2, solve y^2-y+1=0 using the well known formula to obtain two complex solution for y: y=(1±i√3)/2. Take ±√y to obtain the four complex solutions for x: [(√3 + 1)/2, √3 - 1)/2, -√3 + 1)/2, -√3 - 1)/2].
The second solution for the equation 2^a+4^b+8^c=584 is a=9, b=3, c=1. 584 have also 512, 64, and 8 as component of the sum. 512 and 8 can be expressed as 2^9 resp. 8^1 or 8^3 resp.2^3. 64 remains in both solutions 4^3.
Utterly trivial: convert 584 to binary 1001001000, so 584=2^9+2^6+2^3=(2^3)^3+(2^2)^3+2^3=8^3+4^3+2^3.
Solution: a=b=c=3.
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Falta outra possibilidade: a = 9 ; b = 3 e c = 1. ( Another possibility is missing: a = 9 ; b = 3 and c = 1.)
Your Video was very helpful and very nice for education.
But I have a problem in some questions and I am trying hard to hard but not able to find exact answers of variables. Can you solve those for me ?
Q. 1 Given -
a - b = 2 , a^5 - b^5 = 992
a = ? , b = ?
Q. 2. Given -
25x^2 + 49y^2 = 841 ,
xy = 12 , x = ? , y = ?
Q. 3. Given -
(x^2 - 3x)^2 - 8(x^2 - 3x) = 20 , x = ?
Q. 4. Given -
x^4 - x^2 = -1 , x = ?
Please , Solve these questions and help me . Solve when ( or if ) you are free .
Q. 1 Given -
a - b = 2 , a^5 - b^5 = 992
a = ? , b = ?
From the symmetry of the given equations it follows that if (a,b)=(x,y) is a solution, then (a,b)=(-y,-x) is also a solution.
From the first equation, b=a-2, so substitute a-2 for b in the first equation: a^5-(a-2)^5=992. Expand and simplify to obtain a^4-4a^3+8a^2-8a-96=0. f(a)=a^4-4a^3+8a^2-8a-96 is a convex function (the 2nd derivative is >0 for all a), so there can be at most two real solution for a. a=-2 is such a solution, and from the first equation we have b=-4. From the symmetry of the equation we have that a=4, b=2 is the second real solution. If you want the complex solutions, simply divide f(a) by (a+2)(a-4) to obtain a quadratic equation for the remaining complex solutions and solve using the well known formula for quadratic equations.
Q. 2. Given -
25x^2 + 49y^2 = 841 ,
xy = 12 , x = ? , y = ?
y^2=144/x^2 from the second equation. Use this in the first equation and multiply by x^2 to obtain a quadratic equation for z=x^2: 25z^2-841z+49*144. Use the well known formula to obtain z=[841±√(841^2-4*49*144*25)]/50. This simplifies by elementary arithmetic to z=16 or z=441/50. Therefore x=±√z, or x=±4 or x=±21/5. Use the second equation to find the corresponding values for y.
Q. 3. Given -
(x^2 - 3x)^2 - 8(x^2 - 3x) = 20 , x = ?
Let y=(x^2 - 3x). The given equation becomes a simple quadratic equation in y. Use the well known formula bla bla bla to obtain y=-2 or y=10. Then solve the two quadratic equations x^2 - 3x+2=0 and x^2 - 3x+2-10=0 to find the for solutions [1, 2, -2, 5].
Q. 4. Given -
x^4 - x^2 = -1 , x = ?
This has 4 complex and no real solutions. Let y=x^2, solve y^2-y+1=0 using the well known formula to obtain two complex solution for y: y=(1±i√3)/2. Take ±√y to obtain the four complex solutions for x: [(√3 + 1)/2, √3 - 1)/2, -√3 + 1)/2, -√3 - 1)/2].
2^a + 2^(2b) + 2^(3c) = 2^9 + 2^6 + 2^3
a = 3, b = 3, c = 3
C=1