Can a quadratic have one real and one non real root

Mathematics

1 answer:

alukav5142 [94]3 years ago

3 0

Answer:

Yes, it is possible. We already know that complex roots exist in conjugate pair but the case is true only when the equation has real coefficients. But on the other hand, if the equation has complex coefficients, then it is possible to attain one real and one complex root.

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In a reliability test there is a 42% probability that a computer chip survives more than 500 temperature cycles. If a computer c

Strike441 [17]

Answer:

The probability is $P(B' n A ' ) =$ 16%

Step-by-step explanation:

From the question we are told that

A is the event that the chip was "made by company A"

B is the event that a chip "survives 500 temperature cycles"

The probability that a computer chip survives more than 500 temperature cycles is

$P(B) = 42$ % $= 0.42$

The probability that does not survive more than 500 temperature cycles and it was manufactured by company A is $P(A | B' ) = 0.73$

The probability that the computer chip is not manufactured by company A and does not survive more than 500 temperature cycles is mathematically evaluated as

$P(B' n A ' ) = P (B' | A' ) P(B ')$

Where $B' \ and \ A'$ are a A and B complements and | represented AND operator

$P(B' n A ' ) = [1 - P (B' | A )] [1 - P(B )]$

substituting values

$P(B' n A ' ) = [1 - 0.73] [1 - 0.42 ]$

$P(B' n A ' ) = (0.27 )(0.58)$

$P(B' n A ' ) =0.16$

$P(B' n A ' ) =$ 16%

7 0

3 years ago

How to evaluate the limit

anzhelika [568]

$\displaystyle\lim_{x\to2}\frac{x^2-x+6}{x+2}$

Both the numerator and denominator are continuous at $x=2$ , which means the quotient rule for limits applies:

$\dfrac{\displaystyle\lim_{x\to2}(x^2-x+6)}{\displaystyle\lim_{x\to2}(x+2)}=\dfrac{2^2-2+6}{2+2}=\dfrac84=2$

Perhaps you meant to write that $x\to-2$ instead? In that case, you would have

$\displaystyle\lim_{x\to-2}\frac{x^2-x+6}{x+2}=\lim_{x\to-2}\frac{(x+2)(x-3)}{x+2}=\lim_{x\to-2}(x-3)=-2-3=-5$