Answer to Question #247359 in Algebra for M.l

Question

Let f(x)=x^2+bx+c. It has turned out that equation f(x) = 2x - 7 has exactly one solution, and equation f(x)=21-6x also has exactly one solution. Find the largest value of parameter p such that equation f(x) has exactly one solution. 

Answer 1

Given question is missing some parts and parameter p. We assume it as follows:

Let $f(x)=px^2+bx+c$ . It has turned out that equation g(b) = 2b - 7 has exactly one solution, and equation h(c)=21-6c also has exactly one solution. Find the largest value of parameter p such that equation f(x) has exactly one solution. 

Solution:

$g(b)=2b-7=0 \\ \Rightarrow 2b=7 \\ \Rightarrow b=3.5$

$h(c)=21-6c=0 \\\Rightarrow 6c=21 \\ \Rightarrow c=21/6=3.5$

So, $f(x)=px^2+3.5x+3.5$

Now f(x) has exactly one solution, so its discriminant D must be 0.

$D=B^2-4AC=0 \\ \Rightarrow 3.5^2-4p(3.5)=0 \\ \Rightarrow 12.25=14p \\ \Rightarrow p=\dfrac{12.25}{14} \\ \Rightarrow p=0.875$