Because the degree is two, we either have two real roots or two complex roots
The roots are non-real because the discriminant is < 0
2x^2 + 4x + 7 = 0 subtract 7 from both sides
2x^2 + 4x = -7
2 (x ^2 + 2x) = -7 divide both sides by 2
x^2 + 2x = -7/2
Take 1/2 of 2 = 1.....square it = 1 add it to both sides
x^2 + 2x + 1 = -7/2 + 1 factor the left, simplify the right
(x + 1) ^2 = -5/2 take both roots
x + 1 = ± √[ -5/2 ] subtract 1 and simplify the right
x = ± √[5/2] i - 1
Answer:
(a) According to the fundamental theorem of algebra the number of roots for [tex]2x^2+4x+7[/tex] is 2.
(b) The roots of the polynomial are [tex](\dfrac{-4+\sqrt{-40}}{4})(\dfrac{-4-\sqrt{-40}}{4})\\[/tex].
Step-by-step explanation:
Given information:
The given polynomial is [tex]2x^2+4x+7[/tex].
Now, the degree of a polynomial is the highest power of the variable.
So, the given polynomial is a second-degree polynomial which contains two roots.
(a) According to the fundamental theorem of algebra the number of roots for [tex]2x^2+4x+7[/tex] is 2.
(b) Roots of a second degree polynomial can be found as,
[tex]ax^2+bx+c=(x-\dfrac{-b+\sqrt{b^2-4ac}}{2a})(x-\dfrac{-b-\sqrt{b^2-4ac}}{2a})[/tex]
Split the middle terms of the given polynomial as,
[tex]2x^2+4x+7=(x-\dfrac{-4+\sqrt{16-56}}{4})(x-\dfrac{-4-\sqrt{16-56}}{4})\\2x^2+4x+7=(x-\dfrac{-4+\sqrt{-40}}{4})(x-\dfrac{-4-\sqrt{-40}}{4})\\[/tex]
Now, the roots of the polynomial are imaginary because it contains square root of negative number (-40).
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https://brainly.com/question/2925460?referrer=searchResults
