Find the x-intercepts for the parabola defined

Mathematics

1 answer:

sesenic [268]3 years ago

4 0

Answer:

(- 3, 0 ) and (1, 0 )

Step-by-step explanation:

Given

y =2x² + 4x - 6

To find the x- intercepts let y = 0, that is

2x² + 4x - 6 = 0 ← divide all terms by 2

x² + 2x - 3 = 0 ← in standard form

(x + 3)(x - 1) = 0 ← in factored form

Equate each factor to zero and solve for x

x + 3 = 0 ⇒ x = - 3

x - 1 = 0 ⇒ x = 1

The x- intercepts are (- 3, 0 ) and (1, 0 )

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(a + b + c) power of 10 - abc Tip A=1 B=2 C= -3

kicyunya [14]

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$\\ \rm\longmapsto (1+2-3)^{10}-(1)(2)(-3)$

$\\ \rm\longmapsto (3-3)^{10}+6$

$\\ \rm\longmapsto 0^{10}+3$

$\\ \rm\longmapsto 0+3$

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3 years ago

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If a polynomial function f(x) has roots -8, 1, and 6i, what must also be a root of f(x)?

Naddik [55]

Answer:

it must also have the root : - 6i

Step-by-step explanation:

If a polynomial is expressed with real coefficients (which must be the case if it is a function f(x) in the Real coordinate system), then if it has a complex root "a+bi", it must also have for root the conjugate of that complex root.

This is because in order to render a polynomial with Real coefficients, the binomial factor (x - (a+bi)) originated using the complex root would be able to eliminate the imaginary unit, only when multiplied by the binomial factor generated by its conjugate: (x - (a-bi)). This is shown below:

$(x-(a+bi))*(x-(a-bi))=\\(x-a-bi)*(x-a+bi)=\\([x-a]-bi)*([x-a]+bi)=\\(x-a)^2-(bi)^2=\\(x-a)^2-b^2(-1)=\\(x-a)^2+b^2$