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

Answer:

173

Step-by-step 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

Answer:

16.59

Step-by-step explanation:

Given:

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

Expanding the left hand side, we have:

$(ax+b)(bx+a)=abx^2+a^2x+b^2x+ab\\=abx^2+(a^2+b^2)x+ab\\=26x^2+\Box\cdot x+26\\ab=26 \implies b=\frac{26}{a}$

Therefore:

$a^2+b^2=a^2+\dfrac{26}{a} =\dfrac{a^3+26}{a}$

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

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

Recall that:

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

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