2(2y^4 + 7y^3 - 4y^2 -20)
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Y^2
4y^2 - 40/y^2 + 14y - 8
2(-2y^4 - 7y^3 + 4y^2 + 20)
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Y^2
Answer:
1.a=2
2. C x=2 and x=-3
Step-by-step explanation:
The standard form for the quadratic function is
ax^2 +bx+c
so we need to rewrite the function to be in this form
2x^2 -10 = 7x
Subtract 7x from each side
2x^2 -7x-10 = 7x-7x
2x^2 -7x-10 = 0
a =2, b= -7 c=-10
2. The quadratic formula is
-b ± sqrt(b^2 -4ac)
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2a
2x^2 + 2x=12
Lest get the equation in proper form
2x^2 + 2x-12 = 12-12
2x^2 +2x-12 =0
a=2 b=2 c=-12
Lets substitute what we know
-2 ± sqrt(2^2 -4(2)(-12))
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2(2)
-2 ± sqrt(4+96)
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2(2)
-2 ± sqrt(100)
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4
-2 ± 10
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4
-2 + 10 -2-10
----------- and --------------
4 4
8/4 and -12/4
2 and -3
U= -1
-9u+6u=-36+39
-3u=3
u=-1
Move all terms not containing U to the right side of the equation.
Exact Form:
U = 1/2
Decimal Form:
U = 0.5
8y - 6 = 5y + 12
- 5y - 5y
3y - 6 = 12
+ 6 + 6
3y = 18
3 3
y = 6
