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ivolga24 [154]
3 years ago
14

Given the trinomial 2x*2 + 4x + 4, predict the type of solutions. a. Two rational solutions b. One rational solution c. Two irra

tional solutions d. Two complex solutions
Mathematics
2 answers:
Shtirlitz [24]3 years ago
5 0
Given the trinomial 2x2<span> + 4x + 4, predict the type of solutions
</span><span>D = 4² - 4(2)(4) = 16 -32 = -16</span><span>
the determinate is less than 0 meaning 2 complex roots or solutions
D
i got this problem once

</span>
OverLord2011 [107]3 years ago
3 0

Answer:

d. Two complex solutions

Step-by-step explanation:

We have been given a trinomial 2x^2+4x+4 and we are supposed to predict the type of solutions of our given trinomial.

We will use discriminant formula to solve for our given problem.

\text{Discriminant}=b^2-4ac, where,

a =\text{Coefficient of }x^2,

b =\text{Coefficient of }x,

c =\text{Constant }

Conclusion from the result of Discriminant are:

D

D=0\text{ means one real zero with of multiplicity two}

D>0\text{ means two distinct zeroes}

Upon substituting our given values in above formula we will get,

\text{Discriminant}=4^2-4*2*4

\text{Discriminant}=16-32

\text{Discriminant}=-16

Since our discriminant is less than zero, therefore, out given trinomial will have two complex solutions and option d is the correct choice.

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7x + 3x + x⁴ divided by x-2 using long division​
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Step-by-step explanation:

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Total cost for a computer if pay 600 down and 35.00 a month for 12 months
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8 0
3 years ago
Prove that the roots of x2+(1-k)x+k-3=0 are real for all real values of k​
masha68 [24]

Answer:

Roots are not real

Step-by-step explanation:

To prove : The roots of x^2 +(1-k)x+k-3=0x

2

+(1−k)x+k−3=0 are real for all real values of k ?

Solution :

The roots are real when discriminant is greater than equal to zero.

i.e. b^2-4ac\geq 0b

2

−4ac≥0

The quadratic equation x^2 +(1-k)x+k-3=0x

2

+(1−k)x+k−3=0

Here, a=1, b=1-k and c=k-3

Substitute the values,

We find the discriminant,

D=(1-k)^2-4(1)(k-3)D=(1−k)

2

−4(1)(k−3)

D=1+k^2-2k-4k+12D=1+k

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D=k^2-6k+13D=k

2

−6k+13

D=(k-(3+2i))(k+(3+2i))D=(k−(3+2i))(k+(3+2i))

For roots to be real, D ≥ 0

But the roots are imaginary therefore the roots of the given equation are not real for any value of k.

6 0
3 years ago
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