Use the discriminant to determine the number and type of solutions the equation has.

2 answers:

6 0

$ax^2+bx+c=0\ \ \ \Rightarrow\ \ \ \Delta=b^2-4\cdot a\cdot c\\\\\Delta>0\ \ \ \Rightarrow\ \ \ two\ real\ solutions\\\Delta=0\ \ \ \Rightarrow\ \ \ one\ real\ solution\\\Delta0\ \ \ \ \Rightarrow\ \ \ two\ real\ solutions\ \ \ \Rightarrow\ \ \ Ans.\ C$

White raven [17]2 years ago

5 0

$x^2 + 8x + 12 = 0\\ \\a=1 , b= 8, c= 12 \\ \\ \Delta =b^2-4ac = 8^2 -4\cdot1\cdot12= 64-48= 16\\ \\\Delta > 0 \\ \\ Answer : \ C. \ \ two \ rational \ solutions$

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Given

Expenditures at Manager's Store; expenditures at Competitor's Store.

Find

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c) an explanation for the answer in ≥ 2 sentences

Solution

a) The sum of expenditures divided by the number of expenditures (15) is ...

... average for Manager's Store: $37.60

... average for Competitor's Store: $48.53

b) The expenditures at Manager's Store are well-represented by the mean (average).

c) The range of expenditures at Competitor's Store is significantly higher than at Manager's store, so a single number such as mean or median does not represent the data well. The expenditures at Manager's store are more compactly grouped around the mean and median, which are closer together, so the mean is a good representation of Manager's Store expenditures overall.

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2 years ago

Which expression is a fourth root of -1+isqrt3?

aleksklad [387]

Answer:

Step-by-step explanation:

$\sf n^{th}$ roots of a complex number is given by DeMoivre's formula.

$\sf \boxed{\bf r^{\frac{1}{n}}\left[Cos \dfrac{\theta + 2\pi k}{n}+i \ Sin \ \dfrac{\theta+2\pi k}{n}\right]}$

Here, k lies between 0 and (n -1) ; n is the exponent.

$\sf -1 + i\sqrt{3}$

a = -1 and b = √3

$\sf \boxed{r=\sqrt{a^2+b^2}} \ and \ \boxed{\theta = Tan^{-1} \ \dfrac{b}{a}}$

$\sf r = \sqrt{(-1)^2 + 3^2}\\\\ = \sqrt{1+9}\\\\=\sqrt{10}$

$\sf \theta = tan^{-1} \ \dfrac{\sqrt{3}}{-1}\\\\ = tan^{-1} \ (-\sqrt{3})$

$\sf = \dfrac{-\pi }{3}$

n = 4

For k = 0,

$\sf z = \sqrt[4]{10}\left[Cos \ \dfrac{\dfrac{-\pi}{3} +0}{4}+iSin \ \dfrac{\dfrac{-\pi}{3}+0}{4}\right] \\\\\\z= \sqrt[4]{10} \left[Cos \ \dfrac{ -\pi }{12}+iSin \ \dfrac{-\pi}{12}\right]\\\\\\z = \sqrt[4]{10}\left[-Cos \ \dfrac{\pi}{12}-i \ Sin \ \dfrac{\pi}{12}\right]$