sushant singh
Last Activity: 13 Years ago
hey naman,i think this explanation will give you the correct answer...
We have:
√x + y = 11x + √y = 7
If we let a = √x and b = √y, then x = a^2 and y = b^2. This gives the system:
a + b^2 = 11. . . . . . . . . . . .(1)a^2 + b = 7. . . . . . . . . . . . .(2)
From (2), we have:
b = 7 - a^2. . . . . . . . . . . . (3)
Substituting (3) into (2) gives:
a + (7 - a^2)^2 = 11==> a^4 - 14a^2 + a + 49 = 11==> a^4 - 14a^2 + a + 38 = 0==> (a - 2)(a^3 + 2a^2 - 10a - 19) = 0==> a ≈ -3.28, x ≈ -1.85, a = 2, and a ≈ 3.13.
Since a = √x > 0, we see that a = 2 and a ≈ 3.13 are the only solutions by applying this restriction alone. On top of that, we require that b > 0 and then 7 - a^2 = 0 ==> 0 < a < √7. Thus, a = 2 is the only solution valid.
Then:
a = 2 ==> b = 7 - 2^2 = 3.
Therefore, a = 2 and b = 3 ==> x = 4 and y = 9.