A right cylinder with radius 4 cm and height 3 cm
1 answer:
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Answer:
Answer:
System A has 4 real solutions.
System B has 0 real solutions.
System C has 2 real solutions
Step-by-step explanation:
System A:
x^2 + y^2 = 17 eq(1)
y = -1/2x eq(2)
Putting value of y in eq(1)
x^2 +(-1/2x)^2 = 17
x^2 + 1/4x^2 = 17
5x^2/4 -17 =0
Using quadratic formula:
a = 5/4, b =0 and c = -17
Finding value of y:
y = -1/2x
System A has 4 real solutions.
System B
y = x^2 -7x + 10 eq(1)
y = -6x + 5 eq(2)
Putting value of y of eq(2) in eq(1)
-6x + 5 = x^2 -7x + 10
=> x^2 -7x +6x +10 -5 = 0
x^2 -x +5 = 0
Using quadratic formula:
a= 1, b =-1 and c =5
Finding value of y:
y = -6x + 5
y = -6(\frac{1\pm\sqrt{19}i}{2})+5
Since terms containing i are complex numbers, so System B has no real solutions.
System B has 0 real solutions.
System C
y = -2x^2 + 9 eq(1)
8x - y = -17 eq(2)
Putting value of y in eq(2)
8x - (-2x^2+9) = -17
8x +2x^2-9 +17 = 0
2x^2 + 8x + 8 = 0
2x^2 +4x + 4x + 8 = 0
2x (x+2) +4 (x+2) = 0
(x+2)(2x+4) =0
x+2 = 0 and 2x + 4 =0
x = -2 and 2x = -4
x =-2 and x = -2
So, x = -2
Now, finding value of y:
8x - y = -17
8(-2) - y = -17
-16 -y = -17
-y = -17 + 16
-y = -1
y = 1
So, x= -2 and y = 1
System C has 2 real solutions