Question

If (ax+b)(bx+a)=26x^2+ Box(x) +26, where a, b, and Box are distinct integers, what is the minimum possible value of Box, the coefficient of x?

Question in latex: If $(ax+b)(bx+a)=26x^2+\Box\cdot x+26$, where $a$, $b$, and $\Box$ are distinct integers, what is the minimum possible value of $\Box$, the coefficient of $x$?

Answer 1

173

Explanation:

For sympliciy let the box equal y.

Expanding the left side we get (a*x+b)(b*x+a) = (a*b*(x)^2 + (a^2 + b^2)x + a*b). Hence we have that (a*b*(x)^2 + (a^2 + b^2)x + a*b) = 26*(x)^2 + x*y + 26. Scince the coefficients of like terms in our equation must be equal, ab=26. Hence (a,b) = (1,26),(26,1),(-1,-26),(-26,-1),(2,13),(13,2),(-2,-13),(-13,-2). Since a^2 + b^2 = y we can see that the only 2 values of y are 677 and 173 (by simply plug in the values of (a,b)), taking the smaller of the two our answer is [173].

Answer 2

16.59

Explanation:

Given:

`(ax+b)(bx+a)=26x^2+\Box\cdot x+26`

Expanding the left hand side, we have:

`(ax+b)(bx+a)=abx^2+a^2x+b^2x+ab\\=abx^2+(a^2+b^2)x+ab\\=26x^2+\Box\cdot x+26\\ab=26 \implies b=(26)/(a)`

Therefore:

`a^2+b^2=a^2+(26)/(a) =(a^3+26)/(a)`

To find the minimum value, we take the derivative and solve for its critical point.

`(d)/(da) ((a^3+26)/(a))=(2a^3-26)/(a^2)\\$Setting the derivative equal to zero, we have:\\2a^3-26=0\\2a^3=26\\a^3=13\\a=\sqrt[3]{13}`

Recall that:

`\Box=a^2+b^2=\frac{(\sqrt[3]{13}) ^3+26}{\sqrt[3]{13}}\\=\frac{13+26}{\sqrt[3]{13}}\\\\\Box=16.59`

The minimum possible value of the coefficient of x is 16.59.