|
|
|
|
|
|
|
|
|
|
Solution
|
Solution provided by AtoZmath.com
|
|
|
|
Find value of k for which quadratic equation has real roots calculator
|
3. Find value of `k` for which
 1. `x^2-kx-4=0`, has real roots
2. `2x^2+kx+2=0`, has equal roots
3. `mx^2+2x+m=0`, has distinct roots
4. `3x^2+11x+k=0`, has reciprocal roots
5. `x^2+kx+2=0`, has sum of roots = -3
6. `x^2+3x+k=0`, has product of roots = 2
7. `x^2-(k+6)x+2(2k-1)=0`, has sum of roots = 1/2 product of roots
8. `kx^2+2x+3k=0`, has sum of roots = product of roots
9. `x^2+kx+8=0`, has `a-b=2`
10. `x^2-6x+k=0`, has `3a+2b=20`
11. `px^2-14x+8=0`, has `a=6b`
12. `2x^2+kx+2=0`, has one root = 2
|
Example1. Find value of k for which `x^2-kx-4=0` has real roots
Solution:
`x^2-kx-4=0`
Comparing the given equation with `ax^2+bx+c=0`
We get `a=1,b=-k,c=-4`
The equation has real roots.
So, `Delta>=0`
`=>b^2-4ac>=0`
`=>(-k)^2-4*(1)*(-4)>=0`
`=>k^2+16>=0`
`=>k^2>=-16`
|
|
|
|
|
|
|
|
|
|
