Advertisements
Advertisements
प्रश्न
`2x^2+5sqrt3x+6=0`
उत्तर
The given equation is `2x^2+5sqrt3x+6=0`
Comparing it with `ax^2+bx+c=0,` we get
`a=2, b=5sqrt3 and c=6`
∴ Discriminant, `D=b^2-4ac=(5sqrt3)^2-4xx2xx6=75-48=27>0`
So, the given equation has real roots.
Now, `sqrtD=sqrt27=3sqrt3`
∴α=`(-b+sqrtD)/(2a)=(-5sqrt(3)+3sqrt(3))/(2xx2)=(-2sqrt3)/4=-sqrt3/2`
`β=(-b+sqrtD)/(2a)=(-5sqrt(3)+3sqrt(3))/(2xx2)=(-8sqrt3)/4=-2sqrt3`
Hence, `-sqrt3/2`and `-2sqrt3` are the roots of the given equation.
APPEARS IN
संबंधित प्रश्न
Find the value of k for which x = 1is a root of the equation `x^2+kx+3=0`
`3x^2-243=0`
`x^2-6x+4=0`
`15x^2-28=x`
`2x^2-2sqrt2x+1=0`
`3x^2-2sqrt6x+2=0`
`2sqrt3x^2-5x+sqrt3=0`
`12abx^2-(9a^2-8b^2)x-6ab=0,` `Where a≠0 and b≠0`
Find the nature of roots of the following quadratic equations:
`5x^2-4x+1=0`
If -4 is a root of the equation `x^2+2x+4p=0` find the value of k for the which the quadratic equation ` x^2+px(1+3k)+7(3+2k)=0` has equal roots.
