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√x+y=11 and x+√y=7 then what is the value of x and y step to step solution

√x+y=11 and x+√y=7 then what is the value of x and y step to step solution

Grade:12th pass

4 Answers

Chaitya
11 Points
6 years ago
√x + y = 7 x + √y = 11 Let √x = a and √y = b a + b² = 7 ---------- (1) a² + b = 11 ---------- (2) (1) × a² and (2) × a a³ + a²b² = 7a² ---------- (1) a² + ab = 11a ---------- (2) subtract a²b² – ab = 7a² – 11a ab(ab – 1) = a(7a – 11) b(ab – 1) = (7a – 11) ab² – b = 7a – 11 – b = 7a – ab² – 11 – b = a(7 – b²) – 11 a = (11 – b)/(7 – b²) Let b = 1, 2, ... when b = 2 a = (11 – b)/(7 – b²) = (11 – 2)/(7 – 4) = 9/3 = 3 so a = 3 and b = 2 √x = 3 and √y = 2 x = 9 and y = 4
Sachin
13 Points
6 years ago
√x+y=11x+√y=7Let - x=a² and y=b²Now - a+b²=11 a²+b=7=>a+(7-a²)²=11 [∵b=7-a²]=>a+49+a⁴-14a²=11=>a⁴-14a²+a+38=0=>a⁴-2a³+2a³-4a²-10a²+20a-19a+38=0=>a³(a-2)+ 2a²(a-2)-10a (a-2)+ (a-2)
Muhammad mughees zubair
15 Points
5 years ago
√x+y=11 ---(1)and √y +x=7 ----(2)Eq (1) gives √x=11-y→x=(11-y)^2Put in (2) √y+(11-y)^2=7 √y+121+y^2-22y=7 →y^2-22y+√y+114=0--`--(3)Let √y=v → y=v^2 →y^2=v^4So eq. (3) gives v^4-22v^2+v+114=0By synthetic division | 1 0 -22 1 114 | 3 9 -39 -114 3 |___________________ | 3 3 -13 -38| 0So 3 is a rootI.e v=3As √y=v=3 → y=(3)^2→y=9Like wise substituting the value of √y from (2) into (1)We get x=2
Kushagra Madhukar
askIITians Faculty 628 Points
3 years ago
Dear student,
Please find the attached solution to your problem.
√x+y=11 and x+√y=7
Hence, x = (y -11)2
x = y2 -22y + 121
substituting in other equation and assuming √y = z
we get, z4- 22z2+ z + 121 = 7
z = 3 is the only root for which real values of both x and y is obtained
Hence on substituting the value of z we get, y = 9, x = 4
Hope it helps.
Thanks and regards,
Kushagra

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