Find the value of k if x^2+5x+k=0 has one root three times g

[kimmy_koo51]

Find the value of k if x^2+5x+k=0 has one root three times greater than the other.

 

[pka]

\(\displaystyle \L
\begin{array}{l}
x^2 + 5x + k = (x - r)(x - 3r) \\
x^2 + 5x + k = x^2 - 4rx + 3r^2 \\
\Rightarrow \; - 4r = 5\quad \& \quad 3r^2 = k. \\
\end{array}\)

Now solve for k.

 

[soroban]

Re: Find the value of k if x^2+5x+k=0 has one root three tim

Hello, Kimmy!

Here's a back-door approach . . .

Find the value of \(\displaystyle k\) if \(\displaystyle x^2\,+\,5x\,+\,k\:=\:0\) has one root three times greater than the other.

You may not be familiar with this theorem.

If the quadratic \(\displaystyle x^2\,+\,ax\,+b\:=\;0\) has roots \(\displaystyle r\) and \(\displaystyle s\),
. . then: \(\displaystyle \:r\,+\,s\:=\:-a,\;\;rs\:=\:b\)

In baby-talk: .The sum of the roots is the negative of the \(\displaystyle x\)-coefficient;
. . . . . . . . . . .The product of the roots is the constant term.


Let the two roots be \(\displaystyle r\) and \(\displaystyle 3r\)

Then we have: \(\displaystyle \:\begin{array}{cc}r\,+\,3r\:=\:-5\;\; &\Rightarrow & \;\;4r\:=\:-5 \\ r\cdot3r\:=\:k\;\; & \Rightarrow & \;\;3r^2\:=\:k\end{array}\;\;\begin{array}{cc}(1)\\(2)\end{array}\)

From (1), we have: \(\displaystyle \:r\:=\:-\frac{5}{4}\)

Substitute into (2): \(\displaystyle \:k\:=\:3r^2\:=\:3\left(-\frac{5}{4}\right)^2\;\;\Rightarrow\;\;\L k\:=\:\frac{75}{16}\)