I am trying to solve this problem:
If one of the zeroes of the quadratic polynomial $(k-1)x^2+kx+1$ is $-3$, then find $k$.
A) $\frac43$
B) $-\frac43$
C) $\frac23$
D) $-\frac23$
I tried like this: $$a=k-1\qquad b=k\qquad c=1$$ Let $\beta=-3$, as one zero is $-3$.
Now $$\alpha+\beta={-b\over a}={-k\over k-1}$$ and $$\alpha\beta={c\over a}={1\over k-1}$$
Substituting $\beta=-3$, we get
$$\alpha-3={-k\over k-1}$$ and
$$-3\alpha={1\over k-1}\qquad\implies\qquad\alpha={1\over -3(k-1)}=\frac{1}{-3k+1}$$
Substituting in the first equation, we get: $$\frac{1}{-3k+1}-3={-k\over k-1}$$
Solving this, we get $$k = {1\over6} (5-\sqrt{13})\qquad\text{and}\qquad k = {1\over6} (5+\sqrt{13})$$
So where have I gone wrong?
Best Answer
Yet another way: set $x = -3$. Then the equation
$(k - 1)x^2 + kx + 1 = 0 \tag{1}$
becomes
$9(k - 1) -3k + 1 = 0, \tag{2}$
or
$6k - 8 = 0, \tag{3}$
which implies
$k = 4 / 3; \tag{4}$
It looks to me like the answer is (A)!
Hope this helps. Cheerio,
and as always,
Fiat Lux!!!