Algebra 2 question?

Question

how do i solve this?
Find the dimensions of a rectangle whose area is 36 m^2 and whose diagonal is 5 square root of 3 m long.

Answer 1

Let
a, b = dimensions of rectangle

ab = 36
a² + b² = (5√3)² = 75

Solve for b in terms of a.
ab = 36
b = 36/a

Substitute into the second equation.

a² + b² = 75
a² + (36/a)² = 75
a^4 - 75a² + 1296 = 0
(a² - 27)(a² - 48) = 0

a = ±3√3, ±4√3

Since this is a rectangle the negative solutions are rejected.

a = 3√3, 4√3

The rectangle is 3√3 by 4√3.

Answer 2

let the sides be a and b
A= ab = 36 ⇨ b = 36/a

you know that from pythagorean:
a^2 + b^2 = (5√3)^2
a^2 + (36/a)^2 = 75
a^2 + 1296/a^2 = 75
a^4 + 1296 = 75a^2 (multiply everything by a^2)
a^4 - 75a^2 + 1296 = 0

argh, i dont know how to solve quartic roots. =S

but once you solve for a, you can find b.

Answer 3

area is 36.so if l is length and b is breadth then
lb=36 -----(1)
also diagonal is 5*sq root of 3.
so by using pythagoras theorem
l^2+b^2=(5*sq root of 3)^2
l^2+b^2=75
from (1) b=36/l
so l^2+(36/l)^2=75
=>l^2+1296/l^2=75
=>l^4 -75(l^2)+1296=0
l^4 - 48(l^2)-27(l^2)+1296=0.....................
I think u will understand.

Answer 4

For increasing cubic binomials the final formulation is as follows: (a + b) ^ 3 = a^3 + 3*a^2*b^a million + 3*a^a million*b^2 + b^3 on your case, a is x and b is -y^5 So (x - y^5)^3 = x^3 + 3*x^2*(-y^5)^a million + 3*x^a million*(-y^5)^2 + (-y^5)^3 Simplified: =x^3 - 3x^2*y^5 + 3x*y^10 - y^15 :D

Answer 5

area=(l*b)=36sqm
=>l=36/b
diagonal=sqrt(l^2+b^2)=5sqrt3
square the above expression

=>l^2+b^2=25*3=75
=>(36/b)^2+b^2=75
solve for b
and l = 36/b
u will get it.............good luck