PLEASE HELP .!!! ILL GIVE BRAINLIEST.. *EXRTA POINTS* .. DONT SKIP :(( !
1 answer:
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Step-by-step explanation:
b^2-4b+3=0
b²-3x-b+3=0
b(b-3)-1(b-3)=0
(b-3)(b-1)=0
either
b=3 or b=1
.
2n^2 + 7 = -4n + 5
2n²+4n+7-5=0
2n²+4n+2=0
2(n²+2n+1)=0
(n+1)²=0/2
:.n=-1
.
x - 3x^2 = 5+ 2x - x^2
0=5+ 2x - x^2-x +3x^2
0=5+x+2x²
2x²+x+5=0
comparing above equation with ax²+bx +c we get
a=2
b=1
c=5
x={-b±√(b²-4ac)}/2a ={-1±√(1²-4×2×5)}/2×1
={-1±√-39}/2
Given:
The system of equations is
To find:
The graph of the given system of equations and its solution.
Solution:
The table of values for is
x y
-4 7
-2 2
0 -3
Plot these points and connect them by a straight line.
The table of values for is
x y
-4 0
-2 2
0 4
Plot these points and connect them by a straight line.
From the below graph it is clear that the lines intersect each other at point (-2,2).
Therefore, the solution of given system of equation is (-2,2).
