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If x,yx, y are real numbers and equations x212x+35=0x^2 - 12x + 35 = 0 and x2+ax+105=0x^2 + ax + 105 = 0 have at least one common root, then the minimum possible value of y2+4y5ay^2 + 4y - 5a is ___

Entered answer:

Solution

Correct Answer: 106

Solve x212x+35=0x^2 - 12x + 35 = 0: factoring as (x5)(x7)=0(x - 5)(x - 7) = 0, roots are 55 and 77.


The common root must be either 55 or 77.

If 55 is a root of x2+ax+105=0x^2 + ax + 105 = 0: 25+5a+105=025 + 5a + 105 = 0, giving a=26a = -26.

If 77 is a root: 49+7a+105=049 + 7a + 105 = 0, giving a=22a = -22.


For y2+4y5ay^2 + 4y - 5a, complete the square in yy:

y2+4y5a=(y+2)245ay^2 + 4y - 5a = (y + 2)^2 - 4 - 5a

The minimum over real yy is 45a-4 - 5a (when y=2y = -2).


To make 45a-4 - 5a as small as possible, choose the larger value of aa:

For a=22a = -22: 45(22)=4+110=106-4 - 5(-22) = -4 + 110 = 106.

For a=26a = -26: 45(26)=4+130=126-4 - 5(-26) = -4 + 130 = 126.

Minimum possible value =106= 106.

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