Determine the number and type of complex solutions and possible real solutions for each of the following equations. 1) 2x^2 + 5x + 3 = 0 2) 4x^3 – 12x + 9 = 0 3) 2x^4 + x^2 – x + 6 = 0
@Michele_Laino
@mathmale
@robtobey
I have no idea where to even start
@Hero
"1) 2x^2 + 5x + 3 = 0 " is a quadratic equation in the form ax^2 + bx + c=0. By comparing these two equations, determine the values of a, b and c. a= b= c = Now calculate the value of the "discriminant." This is defined as follows:\[discriminant=b^2-4ac\]
a =2 b=5 c=3
25-24=1
@mathmale
@mathmale
Sorry for the delay in my response. You did well in calculating the discriminant. Here are the rules: 1. If the discriminant is positive, you have two real, different roots (e. g., x=4 and x=5) 2. If the discriminant is zero, you have two real, equal roots (e. g., x=2, x=2) 3. If the disc. is negative, you have two complex roots (e. g., x=2+i, x=2-i) Which one of these cases do you have in this problem? How would you describe the roots of your quadratic equation?
there are two real roots
now how would i find the roots? @mathmale
@Michele_Laino
for the first equation, it is suffice to apply the sandard formula: \[\Large x = \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]
Yes. Note, @18jonea, that you have already found b^2 - 4ac: it is 1. So, subst. 5 for b in Michele's equation, above, and 2 for a. then evaluate the two roots. Please share your work here.
\[x=\frac{ -b \pm \sqrt{1} }{ 2a }\]
\[x=-5\sqrt{1}\div4\]
Of course, simplify Sqrt(1) to just plain 1.
-5+1=-4/4=-1
Here, b=4 and a=2. We must write the formula for x using parentheses: x= [-5 plus or minus 1] / 4
-5-1=-6/4= -3/2
Try it, please. What is -5 plus 1? What is -5 minus 1? do this work BEFORE attempting to divide by 4.
Also, after writing one equation, move DOWN (not to the right) to type your next result. Wrong: -5-1=-6/4= -3/2 Right: (-5-1)/4 = -6/4 = -3/2 (answer)
The roots are -1 and -3/2
is that correct?
yes! that's right!
Nice work, 18-J!
thank you, Michele!
:)
Ok Thank you so much could y'all help me with the other two as well please?
I know you have 2 more problems. Please start a new post for each, OK? Thanks.
Ok
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