Yahoo Answers: Answers and Comments for Show that the following equation has real roots for all values of k ? [Mathematics]
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From Anonymous
enUS
Wed, 26 Jan 2011 01:54:23 +0000
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Yahoo Answers: Answers and Comments for Show that the following equation has real roots for all values of k ? [Mathematics]
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From Matt: use the quadratic equation: (b±√(b^24ac))/2a...
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Wed, 26 Jan 2011 10:12:21 +0000
use the quadratic equation: (b±√(b^24ac))/2a where a=(k1), b=2, c=(k3)
(2±√(44(k1)(k3)))2(k1)
Factor out a 4 and multiply out under the square root sign:
(2±√(4(1+k^24k+3)))/2(k1)
Now, square root of 4 is 2, and putting together like terms you have:
(2±2√(k^24k+4))/2(k+1)
Now looking at just what is under the square root sign: k^24k+4, which will factor to (k2)^2.
square root of a square, so you have (2±2(k2))/2(k1)
Factor out a 2: (1±(k2))/(k1)
So, since we were able to factor out of the square root sign without getting a negative number, there can be only real roots.
BUT at the end you have a (k1) in the denominator. so if k=1, we do not have an answer. BUT now look at the original equation. When k=1, the x^2 term becomes 0 and we are left with 2x(13)=0, or 2x+2=0 or x=1. So even when k=1, we still have a real root.
Therefore for all values of k, we will have real roots.

From Fazaldin A: (k1)x^2 + 2x  (k3) = 0
Discriminator = ...
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Wed, 26 Jan 2011 10:12:05 +0000
(k1)x^2 + 2x  (k3) = 0
Discriminator = 2^2 + 4(k1)(k3) =
= 4+4k^216k+12 = 4(k*2 4k +4) =
= 4(k 2)^2 ....................................... [1]
Analysing [1], the equation has real roots for all values of k.
<============ A . N . S . W . E . R ================>

From ?: (k1)x^2 + 2x  (k3) = 0
quadratic equatio...
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https://answers.yahoo.com/question/index?qid=20110126015422AAlSa8T
Wed, 26 Jan 2011 10:08:03 +0000
(k1)x^2 + 2x  (k3) = 0
quadratic equation says that for general equation ax^2+bx+c=0
solution is
x = [b +/ sqrt(b^2  4ac)]/2a
b^2  4ac is the important part to determine the nature of the roots (known as the discriminant)
In our case we can see that the discriminant, D = 4 + 4(k1)(k3) = 4(k2)^2
Therefore, we can now see that for all values of k, D will be a positive real number. ie all roots are real.

From vahucel: If a quadratic equation has real roots, the th...
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Wed, 26 Jan 2011 10:04:01 +0000
If a quadratic equation has real roots, the the discriminant must be = 0 or > 0 ...;
The discriminant is b^2  4ac
Then in your case b^2 4ac = 2^2 4(k1)[(k3)] ... and for real roots
4 +4(k1)(k3)>= 0 ==> 4 + 4(k^2 +3k + 3)>=0 here k^2 +3k +3 has no real root, that is the graph does not cut the x axis... it is positive for all value of K ... the the discriminant is always positive for all value of K. OK!

From Musarrat: (k1)x^2+2x(k3)=0
suppose k1=a and k3=b
...
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Wed, 26 Jan 2011 09:56:16 +0000
(k1)x^2+2x(k3)=0
suppose k1=a and k3=b
so,ab=k1k+3=2
so the equation becomes ax^2+(ab)xb=0
=> ax^2+axbxb=0
=>ax(x+1)b(x+1)=0
=>(axb)(x1)=0
x=1 or x=b/a=(k3)/(k1)