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Firdavs [7]
3 years ago
5

Consider the quadratic equation ax^2+bx+5=0,where a, b and c are rational numbers and the quadratic has two distinct zeros.

Mathematics
1 answer:
FrozenT [24]3 years ago
7 0
You have shared the situation (problem), except for the directions:  What are you supposed to do here?  I can only make a educated guesses.  See below:

Note that if    <span>ax^2+bx+5=0    then it appears that c = 5 (a rational number).

Note that for simplicity's sake, we need to assume that the "two distinct zeros" are real numbers, not imaginary or complex numbers.  If this is the case, then the discriminant,    b^2 - 4(a)(c), must be positive.  Since c = 5, 

b^2 - 4(a)(5) > 0, or b^2 - 20a > 0.

Note that if the quadratic has two distinct zeros, which we'll call "d" and "e," then 

(x-d) and (x-e) are factors of ax^2 + bx + 5 = 0, and that because of this fact,

         - b plus sqrt( b^2 - 20a )
d =  ------------------------------------
                      2a

and

 </span>         - b minus sqrt( b^2 - 20a )
e =  ------------------------------------
                      2a

Some (or perhaps all) of these facts may help us find the values of "a" and "b."  Before going into that, however, I'm asking you to share the rest of the problem statement.  What, specificallyi, were you asked to do here?

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Please Help , Thank You
olya-2409 [2.1K]
I'll go with A and B 

Hope this helps 
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Which of the following points are more than 5 units from the point P(−2, −2)? Select all that apply. A A (2, 1) B B (4, −1) C C
netineya [11]

The distance between any 2 points P(a,b) and Q(c,d) in the coordinate plane, is given by the formula:<span>

<span> |PQ|= \sqrt{ (a-c)^{2} + (b-d)^{2}}</span></span>


Using this formula we calculate the distances |PA|, |PB|, |PC|, |PD| and |PE| and compare to 5.


|PA|= \sqrt{ (-2-2)^{2} + (-2-1)^{2}}= \sqrt{16+9}= \sqrt{25}=5

|PB|= \sqrt{ (-2-4)^{2} + (-2+1)^{2}}= \sqrt{36+1}= \sqrt{37} \approx 6

|PC|= \sqrt{ (-2-2)^{2} + (-2+3)^{2}}= \sqrt{16+1}= \sqrt{17}\approx4

|PD|= \sqrt{ (-2+6)^{2} + (-2+6)^{2}}= \sqrt{16+16}= \sqrt{32}\ \textgreater \  \sqrt{25}=5

|PE|= \sqrt{ (-2+5)^{2} + (-2-1)^{2}}= \sqrt{9+9}= \sqrt{16}=4


Answer: B and D





3 0
3 years ago
The measure of angle 3 is 42°. What is the measure of angle 1 in degrees?
netineya [11]

Answer:

measure of angle 1

= 180° - 90° - 42°

= 48°

7 0
3 years ago
5(a – 1) – 15 = 3(a + 2) + 4
steposvetlana [31]

Answer:

A=15

All you have to do is copy paste your question to the search engine and G0ogle (your search browser) automatically answers it.

3 0
2 years ago
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