Question
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If \( (x+y)^{2}=36, x y=9 \), then \( x^{2}+y^{2}=\cdots \) \( \begin{array}{ll}\text { (a) } 4 & \text { (b) } 27\end{array} \)

Ask by Wells Cross. in Egypt
Mar 10,2025

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Tutor-Verified Answer

Answer

\( x^2 + y^2 = 18 \)

Solution

To find \( x^2 + y^2 \) given the equations \( (x+y)^2 = 36 \) and \( xy = 9 \), we can use the identity: \[ x^2 + y^2 = (x+y)^2 - 2xy \] ### Step 1: Calculate \( x+y \) From the equation \( (x+y)^2 = 36 \), we can find \( x+y \): \[ x+y = \sqrt{36} = 6 \quad \text{(considering the positive root)} \] ### Step 2: Substitute into the identity Now we can substitute \( x+y \) and \( xy \) into the identity: \[ x^2 + y^2 = (x+y)^2 - 2xy \] Substituting the known values: \[ x^2 + y^2 = 36 - 2 \cdot 9 \] ### Step 3: Calculate \( x^2 + y^2 \) Now, calculate: \[ x^2 + y^2 = 36 - 18 = 18 \] ### Conclusion The value of \( x^2 + y^2 \) is \( 18 \). However, this value is not listed in the options provided. Let's double-check the calculations: 1. \( (x+y)^2 = 36 \) gives \( x+y = 6 \). 2. \( xy = 9 \). 3. Using the identity: \[ x^2 + y^2 = 36 - 18 = 18 \] Since \( 18 \) is not among the options (a) 4 or (b) 27, it seems there might be a misunderstanding in the options provided. Thus, the final answer is: \[ \boxed{18} \]

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To solve for \( x^2 + y^2 \), we can use the identity: \[ (x+y)^2 = x^2 + y^2 + 2xy \] From the given information, we have \( (x+y)^2 = 36 \) and \( xy = 9 \). Now substituting these values into the identity gives: \[ 36 = x^2 + y^2 + 2(9) \] Simplifying this: \[ 36 = x^2 + y^2 + 18 \] Next, we subtract 18 from both sides: \[ x^2 + y^2 = 36 - 18 = 18 \] So, the answer is \( x^2 + y^2 = 18 \). Since this answer doesn't match any of the options given, let's check if there's a misunderstanding or if any additional steps can reveal relevant information, but under the current calculations, the value for \( x^2 + y^2 \) is definitely 18. Thus, the final conclusion is that given values do not align with the available options \( (4) \) and \( (27) \).

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