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Maths Question 11 – AP-EAMCET 2026

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If α,β are the roots of the equation x2−15x+1=0, then (1α−15)−2+(1β−15)−2=

Recall the relationships between the roots and coefficients of a quadratic equation.

Step 1: Identify Roots-Coefficients Relations✦ Active

For a quadratic equation ax2+bx+c=0, if α and β are the roots, then the sum of roots α+β=−ba and the product of roots αβ=ca. Given the equation x2−15x+1=0, we can find these relations.

α+β=−(−15)/1=15 αβ=1/1=1
Step 2: Simplify Terms using Root Property○ Expand

Since α is a root of x2−15x+1=0, it satisfies the equation. We can substitute x=α into the equation and manipulate it to find an expression for (1α−15). Since αβ=1e0, it implies that αe0 and βe0, so we can divide by α or β.

α2−15α+1=0 Divide by α: α−15+1α=0 1α−15=−α

Similarly, for the root β:

1β−15=−β
💡 Teacher's Secret Hint

Remember that if the product of roots αβ is non-zero, then individual roots α and β must also be non-zero, allowing division by α and β.

Step 3: Substitute and Rewrite the Expression○ Expand

Substitute the simplified terms back into the original expression that needs to be evaluated.

(1α−15)−2+(1β−15)−2 =(−α)−2+(−β)−2 =1(−α)2+1(−β)2 =1α2+1β2 =β2+α2(αβ)2
Step 4: Calculate α2+β2○ Expand

Use the algebraic identity α2+β2=(α+β)2−2αβ and the values from Step 1.

α2+β2=(15)2−2(1) =225−2=223
Step 5: Final Calculation○ Expand

Substitute the calculated values into the expression derived in Step 3.

α2+β2(αβ)2=223(1)2=223
💡 Teacher's Secret Hint

Double-check all substitutions to avoid calculation errors. A small mistake can lead to a wrong option being selected.

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