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3 years ago

Find value of x for 7(3x+4)+2=156

Mathematics

2 answers:

mr_godi [17]3 years ago

4 0

21x + 28 = 156 - 2
21x = 154 - 28
21x/21 = 126/21
X = 6

liberstina [14]3 years ago

3 0

Answer:

Step-by-step explanation:

21x + 28 + 2 = 156

= 21x = 126

= x = 6

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3 years ago

Solve |x – 4| + 6 = 17.

const2013 [10]

Answer:

$\mathrm{C)\:}x=15,\:-7$

Step-by-step explanation:

For $|a-b|=c$ , there are two possible cases:

$\begin{cases}a-b=c\\a-b=-c\\\end{cases}$

Solving for both cases, we get:

$\begin{cases}x-4+6=17, \:x+2=17, \: \fbox{$x=15$}\\-(x-4)+6=17,\: -x+10=17, \: -x=7, \:, \fbox{$x=-7$}\\\end{cases}$

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2 years ago

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3 years ago

What are the solutions to the equation

frosja888 [35]

Answer:

$x_1=\frac{1}{4}+(\frac{\sqrt{7}}{4})i$ and $x_2=\frac{1}{4}-(\frac{\sqrt{7} }{4})i$

Step-by-step explanation:

You have the quadratic function $2x^2-x+1=0$ to find the solutions for this equation we are going to use Bhaskara's Formula.

For the quadratic functions $ax^2+bx+c=0$ with $a\neq 0$ the Bhaskara's Formula is:

$x_1=\frac{-b+\sqrt{b^2-4.a.c} }{2.a}$

$x_2=\frac{-b-\sqrt{b^2-4.a.c} }{2.a}$

It usually has two solutions.

Then we have $2x^2-x+1=0$ where a=2, b=-1 and c=1. Applying the formula:

$x_1=\frac{-b+\sqrt{b^2-4.a.c} }{2.a}\\\\x_1=\frac{-(-1)+\sqrt{(-1)^2-4.2.1} }{2.2}\\\\x_1=\frac{1+\sqrt{1-8} }{4}\\\\x_1=\frac{1+\sqrt{-7} }{4}\\\\x_1=\frac{1+\sqrt{(-1).7} }{4}\\x_1=\frac{1+\sqrt{-1}.\sqrt{7}}{4}$

Observation: $\sqrt{-1}=i$

$x_1=\frac{1+\sqrt{-1}.\sqrt{7}}{4}\\\\x_1=\frac{1+i.\sqrt{7}}{4}\\\\x_1=\frac{1}{4}+(\frac{\sqrt{7}}{4})i$

And,

$x_2=\frac{-b-\sqrt{b^2-4.a.c} }{2.a}\\\\x_2=\frac{-(-1)-\sqrt{(-1)^2-4.2.1} }{2.2}\\\\x_2=\frac{1-i.\sqrt{7} }{4}\\\\x_2=\frac{1}{4}-(\frac{\sqrt{7}}{4})i$

Then the correct answer is option C.

$x_1=\frac{1}{4}+(\frac{\sqrt{7}}{4})i$ and $x_2=\frac{1}{4}-(\frac{\sqrt{7} }{4})i$

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