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If \(x\sqrt{x} + y\sqrt{y} = 32\) and \(x\sqrt{y} + y\sqrt{x} = 31\), then what is the value of \(x + y\) ?
A. \(\sqrt{5}\)
B. \(5\)
C. \(6.2\)
D. \(12.6\)
E. \(25\)
A. \(\sqrt{5}\)
B. \(5\)
C. \(6.2\)
D. \(12.6\)
E. \(25\)
19 Aug 2022, 02:13
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Official Solution:
If \(x\sqrt{x} + y\sqrt{y} = 32\) and \(x\sqrt{y} + y\sqrt{x} = 31\), then what is the value of \(x + y\) ?
A. \(\sqrt{5}\)
B. \(5\)
C. \(6.2\)
D. \(12.6\)
E. \(25\)
Step 1:
Let \(\sqrt{x}=m\) and \(\sqrt{y} =n\). Notice that in this case, we'd need to find the value of \(x + y=m^2+n^2\).
Then we'd have:
(i) \(m^3 + n^3 = 32\);
(ii) \(mn^2 + nm^2 = 31\)
Step 2:
Recall that:
\((m + n)^3 = m^3 + n^3 + 3mn^2 + 3nm^2\)
Substitute (i) and (ii) in the above:
\((m + n)^3 = 32 + 3*31\)
Take the cube root:
\(m + n = 5\)
Step 3:
From (ii) we have that \(mn(n + m) = 31\);
Substitute \(m + n = 5\) in the above:
\(mn = \frac{31}{5}\)
Step 4:
Recall that \((m + n)^2 = m^2 + n^2 + 2mn\);
Substitute \(m + n = 5\) and \(mn =\frac{31}{5}\) in the above:
\(5^2= m^2 + n^2 + 2*\frac{31}{5}\);
\(m^2 + n^2 = 12.6\)
Answer: D
If \(x\sqrt{x} + y\sqrt{y} = 32\) and \(x\sqrt{y} + y\sqrt{x} = 31\), then what is the value of \(x + y\) ?
A. \(\sqrt{5}\)
B. \(5\)
C. \(6.2\)
D. \(12.6\)
E. \(25\)
Step 1:
Let \(\sqrt{x}=m\) and \(\sqrt{y} =n\). Notice that in this case, we'd need to find the value of \(x + y=m^2+n^2\).
Then we'd have:
(i) \(m^3 + n^3 = 32\);
(ii) \(mn^2 + nm^2 = 31\)
Step 2:
Recall that:
\((m + n)^3 = m^3 + n^3 + 3mn^2 + 3nm^2\)
Substitute (i) and (ii) in the above:
\((m + n)^3 = 32 + 3*31\)
Take the cube root:
\(m + n = 5\)
Step 3:
From (ii) we have that \(mn(n + m) = 31\);
Substitute \(m + n = 5\) in the above:
\(mn = \frac{31}{5}\)
Step 4:
Recall that \((m + n)^2 = m^2 + n^2 + 2mn\);
Substitute \(m + n = 5\) and \(mn =\frac{31}{5}\) in the above:
\(5^2= m^2 + n^2 + 2*\frac{31}{5}\);
\(m^2 + n^2 = 12.6\)
Answer: D
