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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?

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You worked much too hard. Plug in. We get $9(k-1)-3k+1=0$. Solve. – André Nicolas Dec 2 '13 at 6:45
up vote 1 down vote accepted

Yet another way: set $x = -3$. Then the equation

$(k - 1)x^2 + kx + 1 = 0 \tag{1}$


$9(k - 1) -3k + 1 = 0, \tag{2}$


$6k - 8 = 0, \tag{3}$

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!!!

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Thanks. I never thought of this method. – Kartik Dec 2 '13 at 6:50
@ Kartik: my pleasure, sir! And thanks for the "acceptance"! – Robert Lewis Dec 2 '13 at 6:51


Another way:


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I did not understand this. Could you please explain a little more? – Kartik Dec 2 '13 at 6:42
@Kartik, so one of the zeros is $-1,$ what is the other? – lab bhattacharjee Dec 2 '13 at 6:42
Still not understood. One zero is $-1$, other is $-3$, so polynomial should be $(x+1)(x+3)=x^2+4x+3$, so $k=4$, but that is also not in the options? – Kartik Dec 2 '13 at 6:47
@Kartik, $ax^2+bx+c=0$ and $Ax^2+Bx+C=0$ iff $$\frac aA=\frac bB=\frac cC$$. Btw, the roots of $(k−1)x^2+kx+1$ are $-1$ and $\frac{-1}{k-1}$ which needs to be $-3$ right? – lab bhattacharjee Dec 2 '13 at 6:49
Thanks! Now I understood this. – Kartik Dec 2 '13 at 6:52

Simply put x = -3 in your equation $(k-1)*(-3)^2 + k*-3 + 1 = 0$

Solving it, you get k = 4/3

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Thanks. I never thought of this method. – Kartik Dec 2 '13 at 6:50

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