Find the positive value of $n$ such that the equation $9x^2+nx+1=0$ has exactly one solution in $x$.

Guest May 14, 2019

#1**+4 **

Ima reformat da question right quick nawmsayn:

Find the positive value of \(n\) such that the equation \(9x^2+nx+1=0\) has exactly one solution in \(x\).

When working with quadratic equations and the question mentions something about a number of solutions, we definitely want to use the discriminant \(b^2-4ac\).

Let's review what the output of the discriminant tells you about the solutions to a quadratic equation.

If \(b^2-4ac > 0\), there are two solutions.

If \(b^2-4ac = 0\), there is one solution.

If \(b^2-4ac < 0\), there are no solutions.

Since we want only one solution, we want \(b^2-4ac = 0\). Where currently \(a = 9, b = n, c = 1\)

\(b^2 - 4ac = 0 \rightarrow n^2 - 4(9)(1) = 0\rightarrow n^2 = 36 \rightarrow n = \pm 6\)

The question asks for the *positive* value of \(n\), therefore \(n = 6\)

Anthrax May 14, 2019