Solve the equation for all real solutions.

Mathematics

1 answer:

horsena [70]3 years ago

3 0

Can you reformat the question?

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Solve the equation x^2-4x+85=0

lara [203]

Can't factor
use quadratic
for
ax^2+bx+c=0
x= $\frac{-b+/- \sqrt{b^2-4ac} }{2a}$

given
1x^2-4x+85=0
a=1
b=-4
c=85

x= $\frac{-(-4)+/- \sqrt{(-4)^2-4(1)(85)} }{2(1)}$
x= $\frac{4+/- \sqrt{16-340} }{2}$
x= $\frac{4+/- \sqrt{-324} }{2}$
x= $\frac{4+/- (\sqrt{-1})(\sqrt{324}) }{2}$
remember that √-1=i
x= $\frac{4+/- i(\sqrt{324}) }{2}$
x= $\frac{4+/- i(18) }{2}$
x=2+/-9i

x=2+9i or x=2-9i

3 0

3 years ago

Given f(x)2x and g(x) = x^2+3 , find f(g(x)).

MakcuM [25]

Answer:

2x^2 + 6.

Step-by-step explanation:

Replace the x in f(x) by g(x):

f(g(x)) = 2(x^2 + 3)

= 2x^2 + 6.

8 0

3 years ago

A plus 5 equals negative 5a plus 5

jenyasd209 [6]

A+5=5a+5 Subtract a from each side
5=5a+5 Subtract 5 from each side
0=5a Divide each side by the number before the variable to get it alone.
--- ---
5 5
a=0

4 0

3 years ago

Read 2 more answers

I don’t know how to work i this out

snow_tiger [21]

Equation of the line:

y = -0.4x + 1.6

or
y =
- 2 x
5
+ 8
5
When x=0, y = 1.6
When y=0, x = 4

Mark brainliest please

Hope this helps you

8 0

3 years ago

Please help me solve the first one and show all work

Leto [7]

Acc. to midpoint theorem,

15={(2x+9) + (4x-15)}/2
⇒ 30=2x+9+4x-15
⇒ 30-9+15=6x
⇒ 36/6=x
⇒6=x

WZ= 4x-15
=4*6-15
=24-15=9

3 0

3 years ago