try this: a+squareroot of b=41,squareroot of a+b=31.find the values of 'a' and 'b'?

Answer 1

No one has answered it correctly

let a = x^2
b = y^2

x+y^2 = 31 ..1
x^2+y = 41... 2

from 1

x= 31-y^2
so (31-y^2)^2 + y = 41
or 961-62y^2+y^4+y = 41

or y^4-62y^2+y+920 = 0

y if real is a factro or 920

1,2,4,5

by trial and error y =5 makes it zero

y(1) and Y(2) are too high try 4 and 5

so b = 25 and a = 36

this is answer

Answer 2

"sqrt(x)" will represent "the square root of x."

Given:

sqrt(a+b)=31.

Square both sides of the eq'n. Then,

a+b=961
a=961-b

Substitute this value into the first given eq'n:

961 - b + sqrt(b) = 41

A little bit of algebra and:

b - sqrt(b) - 920 = 0

Let sqrt(b) = x. Then,

x^2 - x - 920 = 0.

Use the quadratic formula on the above eq'n (omitted here).

x = (1 +/- 3 sqrt(409)) / 2

However, since x = sqrt(b), x > 0 (I'm assuming 'a' and 'b' are real #s and that you can't have a negative square root). So,

sqrt(b) = 1/2 + 3 sqrt(409) / 2.

b = (3 * sqrt(409) + 1) ^ 2 / 4

Recall that a = 961 - b, so

a = 81/2 - 3 sqrt(409)/2

Not a nice problem...

:-(

Answer 3

1) a+b=31^2=961
2) a+√b=41
2) b=41^2-82a+a^2
1)and 2) a+1681-82a+a^2=961

a^2 - 81a + 720=0

a=70.83 or 50.915 and
b=889.83 or 98.34 respectively

But when a is 50.915 then √98.34 is -9.9
which works for a+√b. But does not work for
√(a+b)=31

Then 70.83 and 889.83 work for √(a+b)=31

a+√b when √889.83=-29.83 So I'll
go with 70.83 and 889.83 for a and b respectively
using the negative square root of b for a+√b

Answer 4

a=36 b=25

Answer 5

The question is worded ambiguously, I assume the system is:

a+sqrt(b)=41 and sqrt(a)+b=31.

If so, the answer set is: a = 36, b = 25.

If, instead, the system is supposed to be

a+sqrt(b)=41 and sqrt(a+b)=31,

then the answer set is:
a = -sqrt(1841/2 + 3/2*sqrt(409)) + 41, b = 1841/2 + 3/2*sqrt(409), or as a decimal, a = 10.16437738, b = 950.8356226

Answer 6

b=(41-a)^2
=1681-82a+a^2
=1681-82a+a^2
a=(31-b)^2
=961-62b+b^2
substituting the value of a in the first equation b can be found and then by substitution a can be found

Answer 7

a+(b)^1/2=41 so a=41-(b)^1/2 substitute in (a)^1/2+b=31 then get roots of b using -b+-(b^2-4ac)^1/2/2a then u get 2 values for b then substitute to get values of a

Answer 8

a*sqrt(b)=41
a^2b=1681


sqrt(a+b)=31
a+b=961
b=961-a
a^2(961-a)=1681
-a^3+961a^2-1681=0
a^3-961a^2+1681=0

There will be at least 1 real root. I don't have the patience to find it. Once you have it, substitute in
b=961-a

Answer 9

a+sqr(b)=41
sqr(a+b)=31

square both side in equation 2:
a+b=961
b=961-a

sub 961-a for b in equation 1
a+sqr(961-a)=41
subtrac a from both sides
sqr(961-a)=41-a
square both sides
961-a=1681-82a+a^2
subtract 961 from both sides
-a=720-82a+a^2
add a from both sides
0=720-81a+a^2

use quad formula to solve for a and sub what you get for a to solve for b

Answer 10

a=36,b=25.