Math help please! Roots of the quadratic equations?
1. For 3x^2 - 24x + 48 = 0, determine the value of the discriminant. What does this value tell you about the equation? How many times does the corresponding function cross the x-axis?
2. For what value of 'k' does the equation x^2 + k = kx - 8 have:
a) two distinct real roots
b) one real root
c) no real root
Please show all work so I can figure out how you got the answer and study from it!! thanks so much.
- 1 decade agoFavorite Answer
ok, firstly you need to know the quadtratic formula. I will try to give it here:
hopefully you've gone through more or less how this formula was derived, but if you haven't then just know that it's a re-arrangement of the general case "ax^2 + bx +c = 0"
for some constants a, b and c
and that it gives 2 solutions for the variable x (that's what the +- sign is for)
now, plugging the values from your equation (3x^2 - 24x + 48 = 0) into our formula
x= -(-24)+sqrt(...) etc etc
the "discriminant" is the value inside the square route... this is (b^2 - 4ac) or (576 - 576) or 0
the fact that this is 0 means that you will add zero when getting one solution, and subtract it for the other... or, that there will only BE one solution... and since there is only one solution, that is how many times (on a graph of the function) you will see the graph touch the x-axis.
2. Firstly, you want to recognize that this is a quadratic equation.
There needs to be a x^2 term, and no higher powers like x^3, x^4 etc...
ok, so, secondly, lets re-write it in the familiar form:
x^2 - kx + k + 8 = 0
that's standard algebraic manipulation
next lets find our values for a, b and c, remember k is just a constant...
a = 1
b = -k
c = k+8
easy enough, now lets solve for x
x = (k+- [sqrt(k^2-4(k+8))])/(2)
yuk, maybe write that down so it makes sense to you... it's just from the formula...
ok, now something interestings happened, our discriminant is yet ANOTHER quadratic equation
namely k^2 - 4k - 32
remember, if our discriminant is larger than 0, we have 2 roots (or solutions, same thing)
if its equal to 0, one solution
and less than 0, we get the square root of a negative, and so no real roots...
so, for a)
discriminant > 0
k^2 - 4k - 32 > 0
(k-8)(k+4) > 0 (factorizing, another basic thing you ought to know how to do)
so k=8 and k=-4 are your points of interest, this is a parabola...
so investigating the parabola I see that x is negative between these points (sub in, say, x = 0)
and thus positive on either side
SO... to answer the question, there are 2 roots when k < - 4 and k > 8
there is one root when k = -4 or k = 8
and there are no real roots when -4 < k < 8
(do you see why this is?)
these questions test a lot of basics with polynomials of the 2nd degree... I suggest you study them well as there is a lot going on...
Good luckSource(s): I am teaching/studying maths.
- zimpaniLv 41 decade ago
For a quadratic equation ax^2 + bx + c = 0 the discriminant is b^2 - 4ac. It tells
you what kind of solutions the equation has, because the solutions are
x = [ -b +/1 sqrt(b^2 - 4ac)] / 2a
where +/- means take the + sign for one case and - sign for the other.
Now, if the discriminant b^2 - 4ac > 0, the square root exists and there are
2 real roots. The curve crosses the x axis twice, once at each root.
If it is exactly zero, there is one solution. The curve touches the x axis at x =
root, but does not cross it.
If discriminant is negative, there is no real root. The sqrt is imaginary, so there
are two complex roots. The curve doesn't cross or touch the X axis.
In question 1, we have a = 3, b = -24, c = 48. Now apply the above wisdom.
- Anonymous4 years ago
If the quadratic ax^2 + bx + c = 0 has roots p, q then p + q = -b/a and pq = c/a. 2x^2 - 6x + 3 = 0 has roots a, b. a + b = -(-6)/2 = 3, ab = 3/2 New sum of roots = a^3 - b + b^3 - a = a^3 + b^3 - (a + b) = (a + b)(a^2 - ab + b^2) - (a + b) = (a + b)((a + b)^2 -3ab) - (a + b) = 3((3^2) - 3*3/2) - 3 = 21/2 New product of roots = (a^3 - b)(b^3 - a) = (ab)^3 + ab - a^3 - b^3 = (ab)^3 + ab - (a^3 + b^3) = (ab)^3 + ab - (a + b)((a + b)^3 - 3ab) = (3/2)^3 + 3/2 - 3((3^2) - 3*3/2) = 27/8 + 3/2 - 27/2 = 27/8 + 12/8 - 104/8 = -65/8 New quadratic is x^2 - (sum of roots)x + product of roots = 0 x^2 - 21/2 - 65/8 = 0 8x^2 - 84x - 65 = 0. EDIT. Yes you are right. I obviously did that in too much of a hurry.
- StevenLv 71 decade ago
1. 3x^2 - 24x + 48 = 0
one real root
x^2 + k = kx - 8
(x^2 + 8)/(x-1) =k
c)no real root at x=1