Solve the following simultaneous equations a) 49x – 57y = 172; 57x – 49y = 252 (elimination method) b) 3x + 2y = 29; 5 – y = 18 (substitution method) c) 3x – y = 2; 2x – y = 3 (graphically) d) y + 2x – 19 = 0; 2x – 3y + 3 = 0 (cramer’s rule) About the author Maya
Answer: yuzufsuustua5autstifukludiygxjgdyoiydoyspu Step-by-step explanation: gofyitsiydIdhlzgiziysuosyaoayousl Reply
Step-by-step explanation: Given :- 2x² – 4x + 5 To Find :- 1/α + 1/β (α – β)² Solution :- We know that Sum of zeroes = -b/a Here b = -4 a = 2 -(-4)/2 4/2 2/1 2 Product of zeroes = c/a c = 5 a = 2 5/2 Now 1/α + 1/β = α + β/αβ = 2/(5/2) = 2/5 × 2/1 = 4/5 b) We know that (α – β)² = (α + β)² – 4αβ (2)² – 4(5/2) 4 – 20/2 8 – 20/2 -12/2 -6 Reply
Answer:
yuzufsuustua5autstifukludiygxjgdyoiydoyspu
Step-by-step explanation:
gofyitsiydIdhlzgiziysuosyaoayousl
Step-by-step explanation:
Given :-
2x² – 4x + 5
To Find :-
1/α + 1/β
(α – β)²
Solution :-
We know that
Sum of zeroes = -b/a
Here
b = -4
a = 2
-(-4)/2
4/2
2/1
2
Product of zeroes = c/a
c = 5
a = 2
5/2
Now
1/α + 1/β = α + β/αβ
= 2/(5/2)
= 2/5 × 2/1
= 4/5
b)
We know that
(α – β)² = (α + β)² – 4αβ
(2)² – 4(5/2)
4 – 20/2
8 – 20/2
-12/2
-6