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Illusion [34]
3 years ago
10

The quadratic equation (3k-2)x^2 +12x+3(k+1) =0 has equal roots. Find the two possible values of k.

Mathematics
1 answer:
romanna [79]3 years ago
5 0
x_1=x_2 \Rightarrow \Delta=0\\
\Delta=12^2-4\cdot(3k-2)\cdot3(k+1)\\
\Delta=144-(36k-24)(k+1)\\
\Delta=144-36k^2-36k+24k+24\\
\Delta=-36k^2-12k+168\\
-36k^2-12k+168=0\\
-3k^2-k+14=0\\
-3k^2+6k-7k+14=0\\
-3k(k-2)-7(k-2)=0\\
-(3k+7)(k-2)=0\\
k=-\frac{7}{3} \vee k=2
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