Determine whether the following equations have a solution or not? Justify your answer.?

Question

a)5x2 + 8x + 7 = 0
b)(7)1/2y2 – 6y – 13(7)1/2 = 0
c)2x2 + x – 1 = 0
d)4/3x2 – 2x + 3/4 = 0
e)2x2 + 5x + 5 = 0
f)p2 – 4p + 4 = 0
g)m2 + m + 1 = 0
h)3z2 + z – 1 = 0




I could not make some of the numbers small like they are to be. In each equation at the begining the 2,s are to be small and above the letter. and the two1/2's above both (7) is also to be small above the (7).

Answer 1

the above equations are quadratic and can be solved by the quadratic formula. use the first as an example:
the quadratic equations tates that x =( -b +/- sqrt(b^2 -4ac))/2a

where b is the coefficient of the linear term in x
and a is the coefficient of the squared term
c is the constant

the first equation is 5x^2 +8x +7 = 0
in the above a=5 b=8 and c =7

put in quadratic formula: x= (-8 +/ sqrt(64-140))/2
x= (-8 +/- sqrt(-76))/2
x = (-8 +/- 8.72i) 2
x = -4 +/- 4.36i
x= (-4 + 4.36i) and x= (-4 - 4.36i)

the i comes in because we had to take the square root of a negative number. since that mathematically is imaginary the i simply stands for the square root of -1.

the other problems can be worked in the same manner just remember if you have to take the square root of a negative number you mus introduce i. All quadratic equations can be solved this way but of course an imafginary solution is rather meaningless.

Answer 2

a)5x² + 8x + 7 = 0
delta = 8² - 4.1.7
d = 64 - 28
d = 36
x = (-8 +/- \/36) : 2.5
x' = (-8 + 6):10
x' = -2 : 10 = -0,2
x" = (-8 - 6):10
x" = -14 : 10 = -1,4
x' = -0,2 and x" = -1,4
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b)(7)1/2y² – 6y – 13(7)1/2 = 0
7/2y² - 6y -91/2 = 0
7/2y² - 6y -91/2 = 0
d = -6² - 4(7/2.-91/2)
d = 673
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c)2x² + x – 1 = 0
d = 1² - 4..2.-1
d = 1 +8 = 9
x = (-1 +/- \/9) : 2.2
x' = (-1 + 3) : 4
x' = 2 : 4 = 1/4
x" = (-1-3) : 4
x" = -4/4 = -1
x' = 1/4 or x" = -1
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d)4/3x² – 2x + 3/4 = 0
d = -2² - 4.4/3.3/4
d = 4 -4 = 0
There are two equals R roots.
x = (2 +/- 0) : 2.4/3
x = 2 * 3/8 = 3/4
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e)2x² + 5x + 5 = 0
d = 5² - 4.2.5
d = 25 - 40
d = -15
There is no Real roots because d < 0.
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f)p² – 4p + 4 = 0
d = -4² - 4.1.4
d = 16 -16
d = 0
There are two equals R roots because d = 0
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g)m² + m + 1 = 0
d = 1² - 4.1.1
d = 1-4
d < 0
There is no R roots.
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h)3z² + z – 1 =
d = 1² - 4.3.-1
d = 1 + 12
d = 13
x = (-1 +/- \/13) : 2.3
x' = (-1 + \/13) : 6
x" = (-1 - \/13) : 6
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Answer 3

nicely, once you take advantage of the quadratic formulation, the determinant, (the massive variety less than the sq. root image) determines no matter if the equation could have actual or complicated thoughts. if the determinant is below 0, then it would want to have complicated (imaginary) thoughts. if the determinant is an similar as 0, then it would want to have a million actual answer if the determinant is larger than 0, then it would want to have 2 actual thoughts. the determinant is b^2 - 4ac so, for the first one: sixty 4-4*5*7 sixty 4-one hundred and forty = -seventy six, so, this a possibility have complicated thoughts. yet only be careful with negatives, like in #3 a million-4*2*-a million a million- (-8) = a million+8 = 9, so this a possibility have actual thoughts

Answer 4

every single one of them has a solution.... all numbers on the equations right after the letters are supposed to be exponents


2x2 is supposed to be 2 times x(squared)

4/3 x 2 is 4/3 times x(squared) it could also be 4/(3 times x squared) but in this case, i think the first one is the closest bet.

Answer 5

If we do it for you, it's only hurting you. You'll never learn it for the test.... Do your own homework please.