Answer to Question #147518 in Algebra for Amir

Question

Let us consider a quadratic polynomial f(x) such that equation f(x)=8x-16 has exactly one root, and equation f(x)=2x-4 has exactly one root. Find the maximum value of a discriminant of f(x)

Answer 1

$f(x) = ax^2 + bx + c$

$1.\ ax^2 + bx + c = 8x - 16$

$2. \ ax^2 + bx + c = 2x - 4$

$1. \ ax^2 + x(b -8) + c + 16 = 0$

$2.\ ax^2 + x(b - 2) + c + 4= 0$

$x_1 = x_2 \leftrightarrow D = 0;$

$D = b^2 -4ac;$

$\therefore$

$1. \ (b - 8)^2 - 4a(c +16) = 0$

$2. \ (b - 2)^2 - 4a(c + 4) = 0$

$1. \ b^2 -16b + 64 - 4ac - 64a = 0$

$2. \ b^2 -4b + 4 -4ac - 16a = 0$

Subtracting 2. from 1. yields:

$-12b + 60 -48a = 0$

$b - 5 + 4a = 0$

$b = 5 -4a \ (*)$

From this and 2. :

$(5 - 4a -2)^2 -4ac -16a = 0$

$(3 - 4a)^2 - 4ac -16a = 0$

$9 -24a +16a^2 - 4ac - 16a = 0$

$16a^2 - 40a + 9 - 4ac = 0$

$-16a^2 + 40a - 9 = -4ac$

From this and $(*)$ :

$D_{f(x)} = b^2 - 4ac =$

$= (5 - 4a)^2 - 16a^2 + 40a - 9 =$

$= 25 - 40a + 16a^2 - 16a^2 + 40a - 9 = 16$

Answer: $D_{max} = 16$