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2 years ago

What are the solutions to 3x^2 = 8x — 6?

Mathematics

1 answer:

Bond [772]2 years ago

5 0

Answer:

First, let's move all terms to one side,

$\sf x^2-8x+6=0$

When given this form (ax^2+bx+c=0), there will be two solutions at the end.

1) $\sf x=\frac{-b+\sqrt{b^2-4ab} }{2b}$

2) $\sf x=\frac{-b-\sqrt{b^2-4ab} }{2b}$

where, a=1, b=-8, c=6

$\sf x=\frac{8+\sqrt{(-8)^2-4\times6} }{2} ,\frac{8-\sqrt{(-8)^2-4\times6} }{2}$

Simplify,

$\sf x=\frac{8+2\sqrt{10} }{2} ,\frac{8-2\sqrt{10} }{2}$

Simplify,

$\sf x=4+\sqrt{10}, 4-\sqrt{10}$

Hope it helps!

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What term is 1/1024 in the geometric sequence,-1,1/4,-1/6..?

Trava [24]

Answer:

$\large\boxed{\text{sixth term is equal to}\ \dfrac{1}{1024}}$

Step-by-step explanation:

The explicit formula for a geometric sequence:

$a_n=a_1r^{n-1}$

$a_n$ - n-th term

$a_1$ - first term

$r$ - common ratio

$r=\dfrac{a_2}{a_1}=\dfrac{a_3}{a_2}=...=\dfrac{a_n}{a_{n-1}}$

We have

$a_1=-1,\ a_2=\dfrac{1}{4},\ a_3=-\dfrac{1}{6},\ ...$

The common ratio:

$r=\dfrac{\frac{1}{4}}{-1}=-\dfrac{1}{4}\\\\r=\dfrac{-\frac{1}{6}}{\frac{1}{4}}=-\dfrac{1}{6}\cdot\dfrac{4}{1}=-\dfrac{2}{3}\neq-\dfrac{1}{4}$

<h2>It's not a geometric sequence.</h2>

If $a_3=-\dfrac{1}{16}$ then the common ratio is $r=\dfrac{-\frac{1}{16}}{\frac{1}{4}}=-\dfrac{1}{16}\cdot\dfrac{4}{1}=-\dfrac{1}{4}$

Put to the explicit formula:

$a_n=-1\left(-\dfrac{1}{4}\right)^{n-1}$

Put $a_n=\dfrac{1}{1024}$ and solve for <em>n </em>:

$-1\left(-\dfrac{1}{4}\right)^{n-1}=\dfrac{1}{1024}\qquad\text{use}\ a^n:a^m=a^{n-m}\\\\-\left(-\dfrac{1}{4}\right)^n:\left(-\dfrac{1}{4}\right)^1=\dfrac{1}{1024}\\\\-\left(-\dfrac{1}{4}\right)^n\cdot(-4)=\dfrac{1}{1024}\\\\(4)\left(-\dfrac{1}{4}\right)^n=\dfrac{1}{1024}\qquad\text{divide both sides by 4}\ \text{/multiply both sides by}\ \dfrac{1}{4}/\\\\\left(-\dfrac{1}{4}\right)^n=\dfrac{1}{4096}\\\\\dfrac{(-1)^n}{4^n}=\dfrac{1}{4^6}\qquad n\ \text{must be even number. Therefore}\ (-1)^n=1$