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Lilit [14]
3 years ago
13

Which quadratic equation has no real roots?

Mathematics
2 answers:
Stella [2.4K]3 years ago
6 0

Answer:

The first choice and the last have no real roots.

Step-by-step explanation:

For an equation to have no real roots the value of the discriminant ( b^2 - 4ac) will be negative.

Working out the values:

1.  D = (-2)^2 - 4* -1*-6 = -20 so it has no real roots.

2. D = (3)^2 - 4*  -2* 4 =  41 so it does have real roots.

3. D = (1)^2 - 4* 2*-6 = 49 so real roots.

4. D =   (-1)^2 - 4* 2*3 = -23 so  no real roots

n200080 [17]3 years ago
3 0

Answer:

A and D

Step-by-step explanation:

By finding the roots of all the equations, we see that options A and D are the only ones that require an imaginary number in order to solve

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When the coordinates (1,1) (7,3) (8,0) and (2,-2) are joined which shape is formed ?
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Hope this helps,

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4 0
3 years ago
Can you help me? sin 35 (degrees) Thanks!
Nata [24]
<span>OK, here's a *very* approximate "creative" method: 
</span><span>
sin(a+b)=sin(a)cos(b) + cos(a)sin(b) </span>
<span>
choose a=30deg and b=5deg </span>
- sin(a)=0.5 (exact) 
- b is a small number so we say sin(b)~b (in radians). We can do a long division for this: 3.14/36 = 0.087 radians 
- similarly, for b sufficiently small, cos(b)~1 
<span>- now we only need cos(a) = sqrt(3)/2. We can approximately calculate sqrt(3) as 1.73, (by trial-and-error or better, by using the Babylonian method;  so sqrt(3)/2 = 0.865 </span>
<span>
Finally, sin(30 + 5) = (0.5 x 1) + (0.087 x 0.865) = 0.575 (calculating more digits is of no use because the sin(b)=b and cos(b)=1 are quite inaccurate approximations). </span>

<span>So that wasn't all too much effort, and we're close to the correct answer (0.574)


I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!
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3 years ago
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#9A<br> What is a reasonable first step to solve for y?<br> 1/ 4x + 3y = 2z
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3 years ago
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