Question:
If the equation
\[
x^4-10x^3+37x^2-60x+36=0
\]
has two distinct real roots, where each one of them is a repeated root, then the sum of squares of all the roots of the given equation is
If the equation
\[
x^4-10x^3+37x^2-60x+36=0
\]
has two distinct real roots, where each one of them is a repeated root, then the sum of squares of all the roots of the given equation is
Show Hint
When repeated roots are mentioned in polynomial equations, rewrite the polynomial as powers of linear factors to simplify coefficient comparison.
Updated On: Jun 17, 2026
- $74$
- $26$
- $52$
- $68$
Show Solution
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The Correct Option is B
Solution and Explanation
Concept:
If a quartic polynomial has two repeated roots, then it can be expressed as:
\[
(x-a)^2(x-b)^2
\]
Also, using identities:
\[
a^2+b^2=(a+b)^2-2ab
\]
helps in evaluating the required sum.
Step 1: Assume repeated roots.
Given: \[ x^4-10x^3+37x^2-60x+36=0 \] Since both roots are repeated: \[ =(x-a)^2(x-b)^2 \] \[ =\left[x^2-(a+b)x+ab\right]^2 \] Comparing with: \[ x^4-10x^3+37x^2-60x+36 \] we get: \[ 2(a+b)=10 \] \[ a+b=5 \] Also: \[ (ab)^2=36 \] \[ ab=6 \]
Step 2: Find $a^2+b^2$.
\[ a^2+b^2=(a+b)^2-2ab \] \[ =5^2-2(6) \] \[ =25-12 \] \[ =13 \]
Step 3: Find the sum of squares of all roots.
Since each root repeats twice: \[ a,a,b,b \] Required sum: \[ 2a^2+2b^2 \] \[ =2(a^2+b^2) \] \[ =2(13) \] \[ =26 \] Hence, \[ \boxed{26} \]
Step 1: Assume repeated roots.
Given: \[ x^4-10x^3+37x^2-60x+36=0 \] Since both roots are repeated: \[ =(x-a)^2(x-b)^2 \] \[ =\left[x^2-(a+b)x+ab\right]^2 \] Comparing with: \[ x^4-10x^3+37x^2-60x+36 \] we get: \[ 2(a+b)=10 \] \[ a+b=5 \] Also: \[ (ab)^2=36 \] \[ ab=6 \]
Step 2: Find $a^2+b^2$.
\[ a^2+b^2=(a+b)^2-2ab \] \[ =5^2-2(6) \] \[ =25-12 \] \[ =13 \]
Step 3: Find the sum of squares of all roots.
Since each root repeats twice: \[ a,a,b,b \] Required sum: \[ 2a^2+2b^2 \] \[ =2(a^2+b^2) \] \[ =2(13) \] \[ =26 \] Hence, \[ \boxed{26} \]
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