Given:
[tex]x^2+kx-98=0[/tex]
One of the two roots of the given equation is double the additive inverse of the other.
To find:
The value of k.
Solution:
Let two roots of the given equation are [tex]\alpha[/tex] and [tex]\beta[/tex].
According to the question,
[tex]\beta=2(-\alpha)[/tex] (Additive inverse of [tex]\alpha[/tex] is [tex]-\alpha[/tex])
[tex]\beta=-2\alpha[/tex]
If [tex]\alpha[/tex] and [tex]\beta[/tex] are roots of the quadratic equation [tex]ax^2+bx+c=0,[/tex] then
[tex]\alpha+\beta=-\dfrac{b}{a}[/tex]
[tex]\alpha\beta=\dfrac{c}{a}[/tex]
We have,
[tex]x^2+kx-98=0[/tex]
Here, a =1, b=k and c=-98.
[tex]\alpha+\beta=-\dfrac{k}{1}[/tex]
[tex]\alpha-2\alpha=-k[/tex] [tex][\because \beta=-2\alpha][/tex]
[tex]-\alpha=-k[/tex]
[tex]\alpha=k[/tex] ...(i)
Now,
[tex]\alpha\beta=\dfrac{c}{a}[/tex]
[tex]\alpha(-2\alpha)=\dfrac{-98}{1}[/tex] [tex][\because \beta=-2\alpha][/tex]
[tex]-2\alpha^2=-98[/tex]
Divide both sides by -2.
[tex]\alpha^2=49[/tex]
Taking square root on both sides.
[tex]\alpha=\pm \sqrt{49}[/tex]
[tex]\alpha=\pm 7[/tex]
Using (i), we get
[tex]k=\pm 7[/tex]
Therefore, the value of k is either -1 or 1.
