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Difficulty: Very hardQuadratic Equations

In the quadratic equation x2−ax+b=0x^2 - ax + b = 0, the roots rr and ss are positive integers. The quadratic equation x2−(a+12)x+(b+12)=0x^2 - (a+12)x + (b+12) = 0 has roots r2r^2 and s2s^2, where aa and bb are constants. What is the value of aa?

Answer: 5

Answer

The value of aa is 5.
Applying Vieta's formulas to both equations and using the algebraic identity r2+s2=(r+s)2−2rsr^2 + s^2 = (r+s)^2 - 2rs yields the system of equations a2−2b=a+12a^2 - 2b = a + 12 and b2=b+12b^2 = b + 12. Solving the second equation gives b=4b = 4 (discarding the negative root since r,s>0r, s > 0). Substituting b=4b = 4 into the first equation yields a2−a−20=0a^2 - a - 20 = 0, which solves to a=5a = 5 (discarding a=−4a = -4 since r,s>0r, s > 0).

Step-by-Step Solution

1
Apply Vieta's formulas to the first quadratic equation.
r+s=ar+s = a and rs=brs = b
To express the sum and product of the roots of the first equation in terms of the coefficients aa and bb.
2
Apply Vieta's formulas to the second quadratic equation.
r2+s2=a+12r^2+s^2 = a+12 and r2s2=b+12r^2s^2 = b+12
To express the sum and product of the squared roots in terms of the coefficients of the second equation.
3
Express r2+s2r^2+s^2 in terms of aa and bb using the algebraic expansion of a binomial square.
a2−2b=a+12a^2 - 2b = a+12
By rewriting r2+s2r^2+s^2 as (r+s)2−2rs(r+s)^2 - 2rs, we can substitute aa and bb to form a system of equations.
4
Solve for the constant bb using the product relation.
b=4b = 4
The relation (rs)2=b2=b+12(rs)^2 = b^2 = b+12 simplifies to the quadratic equation b2−b−12=0b^2 - b - 12 = 0. Factoring yields b=4b = 4 or b=−3b = -3. Since rr and ss are positive integers, their product bb must be positive, meaning b=4b = 4.
5
Substitute b=4b = 4 into the sum relation to solve for aa.
a=5a = 5
Substituting b=4b = 4 yields a2−2(4)=a+12a^2 - 2(4) = a+12, which simplifies to a2−a−20=0a^2 - a - 20 = 0. Factoring yields a=5a = 5 or a=−4a = -4. Since rr and ss are positive integers, their sum aa must be positive, meaning a=5a = 5.

Key Concept

Vieta's Formulas and Symmetric Polynomial Relationships
Estimated Time:3m 0s
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Topics

Quadratic Equations

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