Advertisements
Advertisements
Question
For what value of k are the roots of the quadratic equation `kx(x-2sqrt5)+10=0`real and equal.
Solution
The given equation is
`kx(x-2sqrt5)+10=0`
⇒` kx^2-2sqrt5kx+10=0`
This is of the form `ax^2+bx+c=0` where `a=k, b=-2sqrt5k and c=10`
The given equation will have real and equal roots if `D=0`
∴ `20k^2-40k=0`
⇒`20k(k-2)=0`
⇒`k=0 or k-2=0`
⇒` k=0 or k=2`
But, for k=0 we get `10=0` which is not true
Hence, 2 is the required value of k.
RELATED QUESTIONS
If the roots of the equation (a2 + b2) x2 – 2(ac + bd) x + (c2 + d2) = 0 are equal, prove that `a/b = c/d`
In the following, determine whether the given quadratic equation have real roots and if so, find the roots:
3a2x2 + 8abx + 4b2 = 0, a ≠ 0
In the following, determine whether the given quadratic equation have real roots and if so, find the roots:
3x2 - 5x + 2 = 0
`x^2-6x+4=0`
`sqrt3x^2+10x-8sqrt3=0`
`2x^2+5sqrt3x+6=0`
`3a^2x^2+8abx+4b^2=0`
`12abx^2-(9a^2-8b^2)x-6ab=0,` `Where a≠0 and b≠0`
Find the values of k for which the given quadratic equation has real and distinct roots:
`x^2-kx+9=0`
If the roots of the equations `ax^2+2bx+c=0` and `bx^2-2sqrtacx+b=0`are simultaneously real then prove that `b^2=ac`
