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Zorluk: ZorQuadratic Equations and Polynomial Factoring

If the quadratic equation x2kx+36=0x^2 - kx + 36 = 0 has two distinct positive integer roots, r1r_1 and r2r_2, and r1+2r_1 + 2 and r2+2r_2 + 2 are the roots of the quadratic equation x219x+m=0x^2 - 19x + m = 0, what is the value of mm?

Cevap: 70

Cevap

The value of mm is 70.
By applying Vieta's formulas to both quadratic equations, we find that the sum of the original roots is r1+r2=15r_1 + r_2 = 15 and their product is r1r2=36r_1 r_2 = 36. Factoring x215x+36=(x3)(x12)=0x^2 - 15x + 36 = (x-3)(x-12) = 0 confirms the original roots are 33 and 1212. The new roots are 3+2=53+2=5 and 12+2=1412+2=14. Their product m=5×14=70m = 5 \times 14 = 70.

Adım Adım Çözüm

1
Apply Vieta's relations to the first quadratic equation
r1+r2=kr_1 + r_2 = k and r1r2=36r_1 \cdot r_2 = 36
For any quadratic equation x2+bx+c=0x^2 + bx + c = 0, the sum of roots is b-b and the product of roots is cc.
2
Apply Vieta's relations to the second quadratic equation
(r1+2)+(r2+2)=19    r1+r2=15(r_1 + 2) + (r_2 + 2) = 19 \implies r_1 + r_2 = 15
The coefficient of xx in x219x+m=0x^2 - 19x + m = 0 dictates that the sum of its roots equals 1919.
3
Determine the roots r1r_1 and r2r_2 and check integer constraints
r1=3r_1 = 3 and r2=12r_2 = 12
The integer factors of 3636 that sum to 1515 are 33 and 1212, satisfying all conditions.
4
Compute the constant term mm for the transformed equation
m=(3+2)(12+2)=5×14=70m = (3 + 2)(12 + 2) = 5 \times 14 = 70
The constant term mm equals the product of the transformed roots (r1+2)(r_1 + 2) and (r2+2)(r_2 + 2).

Anahtar Kavram

Vieta's Formulas and Quadratic Root Transformations
Tahmini Süre:2m 0s
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