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3 years ago

3√2x+1 as lim = Δx →0

Mathematics

1 answer:

vodomira [7]3 years ago

6 0

1.0

As x -> 0 2x -> 0, SqRt(2x) -> 0 & 3 SqRt(2x) -> 0
Hence:
3 SqRt(2x) + 1 -> 1.0

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Lina20 [59]

Option D: Two irrational solutions

Explanation:

The equation is $17+3 x^{2}=6 x$

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$3x^{2} -6x+17=0$

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$x=\frac{6 \pm \sqrt{36-4(3)(17)}}{2(3)}$

Simplifying the expression, we get,

$\begin{aligned}x &=\frac{6 \pm \sqrt{36-204}}{6} \\&=\frac{6 \pm \sqrt{-168}}{6} \\&=\frac{6 \pm 2 i \sqrt{42}}{6}\end{aligned}$

Taking out the common terms and simplifying, we have,

$\begin{aligned}x &=\frac{2(3 \pm i \sqrt{42})}{6} \\&=\frac{(3 \pm i \sqrt{42})}{3}\end{aligned}$

Dividing by 3, we get,

$x=1+i \sqrt{\frac{14}{3}}, x=1-i \sqrt{\frac{14}{3}}$

Hence, the equation has two irrational solutions.

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