Answer:
x = 3 + √6 ; x = 3 - √6 ; ;
Step-by-step explanation:
Relation given in the question:
(x² − 6x +3)(2x² − 4x − 7) = 0
Now,
for the above relation to be true the following condition must be followed:
Either (x² − 6x +3) = 0 ............(1)
or
(2x² − 4x − 7) = 0 ..........(2)
now considering the equation (1)
(x² − 6x +3) = 0
the roots can be found out as:
for the equation ax² + bx + c = 0
thus,
the roots are
or
or
and, x =
or
and, x =
or
x = 3 + √6 and x = 3 - √6
similarly for (2x² − 4x − 7) = 0.
we have
the roots are
or
or
and, x =
or
and, x =
or
and, x =
or
and,
Hence, the possible roots are
x = 3 + √6 ; x = 3 - √6 ; ;
Answer:
yy
Step-by-step explanation:iii
I believe the answer is 36= -10.4!
Sorry if I'm wrong though.
1) Create was we already know in the equation. 16=-7(-3)+b. (This is using the initial equation of y=mx+b.)
2) Simplify and solve for b: 16=21+b = -5=b
3) Replug your be value into your equation: y=-7x-5.
Answer
y=-7x-5
Words: y equals negative seven x, minus five.