Hi all,
How do I find the value of c for the question below?
The roots of the equation $\displaystyle x^2 + 6x + c = 0$ are k and k-1. Find the value of c.
Solving the quadratic equation
$\displaystyle \displaystyle \begin{align*} x^2 + 6x + c &= 0 \\ x^2 + 6x + 3^2 - 3^2 + c &= 0 \\ (x + 3)^2 + c - 9 &= 0 \\ (x + 3)^2 &= 9 - c \\ x + 3 &= \pm \sqrt{9 - c} \\ x &= -3 \pm \sqrt{9 - c} \end{align*}$
So $\displaystyle \displaystyle k = -3 + \sqrt{9-c}$ and $\displaystyle \displaystyle k - 1 = -3 - \sqrt{9 - c}$.
Solve these for $\displaystyle \displaystyle c$
Hi Prove It,
Thanks for your reply. Am very confused with the solution. why do you have the $\displaystyle 3^2 - 3^2 $
Anyway, I solve for c and I come up with c being 21.25, can you confirm if that is correct? How do I check that the value of c that I come up with is correct?
Hello, bartholomew!
Another approach . . .
$\displaystyle \text{The roots of the equation }\,x^2 + 6x + c \,=\, 0\,\text{ are }k\text{ and }k-1.$
$\displaystyle \text{Find the value of }c.$
Since $\displaystyle x = k$ is a root of the equation,
. . we have: .$\displaystyle k^2 + 6k + c \:=\:0$ .[1]
Since $\displaystyle x=k+1$ is a root of the equation,
. . we have: .$\displaystyle (k-1)^2 + 6(k-1) + c \:=\:0 \quad\Rightarrow\quad k^2 + 4k + c - 5 \:=\:0$ .[2]
Subtract [1] - [2]: .$\displaystyle 2k + 5 \:=\:0 \quad\Rightarrow\quad k \,=\,\text{-}\tfrac{5}{2}$
Substitute into [1]: .$\displaystyle (\text{-}\tfrac{5}{2})^2 + 6(\text{-}\tfrac{5}{2}) + c \:=\:0 \quad\Rightarrow\quad \boxed{c \:=\:\frac{35}{4}}$