+0  
 
0
402
2
avatar

Find the set of values of k for which this equation has two real roots.

5x^2 + 4k*x + k = 0

thanks in advance lol

 Dec 3, 2020
 #1
avatar+128399 
+1

If  this equation  has two real roots,  the discriminant must be >  0

 

Therefore

 

(4k)^2  - 4 (5)(x)   >  0

 

16k^2  - 20k  >   0

 

4k ( 4k  - 5)  >  0

 

One solution is that    k >   5/4

 

The other solution is  k <  0

 

 

cool cool cool

 Dec 3, 2020
 #2
avatar
+1

Well let's get started, the quadratic equation formula is( -b+-sqrt(b^2-4ac))/2, next, we get

a=5

b=4k

c=k

Next we plug it in,[ -4k +- sqrt(16k^2-20k)]/10, now I cannot give the answer because that is not letting you learn, so you must find the value of 16k^2 - 20k that is greater than 0, > 0. Solve the question and good luck 👍 

 Dec 3, 2020

4 Online Users

avatar
avatar
avatar