Question:
Which of the following quadratic equations whose real roots \(x_1,x_2\) satisfy the conditions
\[
x_1^2+x_2^2=5,
\quad
3(x_1^5+x_2^5)=11(x_1^3+x_2^3)?
\]
Which of the following quadratic equations whose real roots \(x_1,x_2\) satisfy the conditions
\[
x_1^2+x_2^2=5,
\quad
3(x_1^5+x_2^5)=11(x_1^3+x_2^3)?
\]
Show Hint
For option-based quadratic root questions, it is often quickest to factorize the options and directly verify the given root conditions.
Updated On: Jun 26, 2026
- \(x^2+3x+2=0\)
- \(x^2+3x+11=0\)
- \(x^2+5x+2=0\)
- \(x^2+5x+11=0\)
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The Correct Option is A
Solution and Explanation
Step 1: Check option (1).
The equation is \[ x^2+3x+2=0 \] Factorizing, \[ x^2+3x+2=(x+1)(x+2) \] So, the roots are \[ x_1=-1,\quad x_2=-2 \]
Step 2: Verify the first condition.
\[ x_1^2+x_2^2=(-1)^2+(-2)^2 \] \[ =1+4 \] \[ =5 \] Thus, the first condition is satisfied.
Step 3: Verify the second condition.
First, \[ x_1^5+x_2^5=(-1)^5+(-2)^5 \] \[ =-1-32 \] \[ =-33 \] So, \[ 3(x_1^5+x_2^5)=3(-33) \] \[ =-99 \] Now, \[ x_1^3+x_2^3=(-1)^3+(-2)^3 \] \[ =-1-8 \] \[ =-9 \] So, \[ 11(x_1^3+x_2^3)=11(-9) \] \[ =-99 \] Hence, \[ 3(x_1^5+x_2^5)=11(x_1^3+x_2^3) \]
Step 4: Final conclusion.
Both given conditions are satisfied by the roots of \[ x^2+3x+2=0 \] Therefore, \[ \boxed{x^2+3x+2=0} \]
The equation is \[ x^2+3x+2=0 \] Factorizing, \[ x^2+3x+2=(x+1)(x+2) \] So, the roots are \[ x_1=-1,\quad x_2=-2 \]
Step 2: Verify the first condition.
\[ x_1^2+x_2^2=(-1)^2+(-2)^2 \] \[ =1+4 \] \[ =5 \] Thus, the first condition is satisfied.
Step 3: Verify the second condition.
First, \[ x_1^5+x_2^5=(-1)^5+(-2)^5 \] \[ =-1-32 \] \[ =-33 \] So, \[ 3(x_1^5+x_2^5)=3(-33) \] \[ =-99 \] Now, \[ x_1^3+x_2^3=(-1)^3+(-2)^3 \] \[ =-1-8 \] \[ =-9 \] So, \[ 11(x_1^3+x_2^3)=11(-9) \] \[ =-99 \] Hence, \[ 3(x_1^5+x_2^5)=11(x_1^3+x_2^3) \]
Step 4: Final conclusion.
Both given conditions are satisfied by the roots of \[ x^2+3x+2=0 \] Therefore, \[ \boxed{x^2+3x+2=0} \]
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