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The equation $3x^2-2\sqrt{6}x+2=0$ has repeated factors after factorisation. What are its roots?
A$\sqrt{\frac{2}{3}}$ and $-\sqrt{\frac{2}{3}}$
B$\sqrt{\frac{3}{2}}$ and $\sqrt{\frac{3}{2}}$
C$\sqrt{\frac{2}{3}}$ and $\sqrt{\frac{2}{3}}$
D$-\sqrt{\frac{2}{3}}$ and $-\sqrt{\frac{2}{3}}$
Answer & Solution
Correct answer: C. $\sqrt{\frac{2}{3}}$ and $\sqrt{\frac{2}{3}}$
We write $3x^2-2\sqrt{6}x+2=(\sqrt{3}x-\sqrt{2})(\sqrt{3}x-\sqrt{2})=(\sqrt{3}x-\sqrt{2})^2$. Thus the same linear factor occurs twice. Solving $\sqrt{3}x-\sqrt{2}=0$ gives $x=\sqrt{\frac{2}{3}}$, so both roots are equal to $\sqrt{\frac{2}{3}}$.
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