1. If `alpha` and `beta` are roots of quadratic equation `3x^2+8x+2=0`, then find the value of `alpha^2+beta^2`
Solution:
`3x^2+8x+2=0`
Comparing the given equation with `ax^2+bx+c=0`
We get `a=3,b=8,c=2`
Sum of roots `=alpha+beta=(-b)/a=(-8)/3`
Product of roots `=alpha*beta=c/a=2/3`
Now we have to find `alpha^2+beta^2`
`alpha^2+beta^2=52/9`
We know that
`alpha^2+beta^2=(alpha+beta)^2-2alphabeta`
`:.alpha^2+beta^2=((-8)/3)^2-2*2/3`
`:.alpha^2+beta^2=64/9-4/3`
`:.alpha^2+beta^2=52/9`
`:.alpha^2+beta^2=52/9`
2. If `alpha` and `beta` are roots of quadratic equation `3x^2+8x+2=0`, then find the value of `alpha-beta`
Solution:
`3x^2+8x+2=0`
Comparing the given equation with `ax^2+bx+c=0`
We get `a=3,b=8,c=2`
Sum of roots `=alpha+beta=(-b)/a=(-8)/3`
Product of roots `=alpha*beta=c/a=2/3`
Now we have to find `alpha-beta`
`alpha-beta=(2sqrt(10))/3`
We know that
`(alpha-beta)^2=(alpha+beta)^2-4alphabeta`
`:.(alpha-beta)^2=((-8)/3)^2-4*2/3`
`:.(alpha-beta)^2=64/9-8/3`
`:.(alpha-beta)^2=40/9`
`:.alpha-beta=(2sqrt(10))/3`
`:.alpha-beta=(2sqrt(10))/3`
3. If `alpha` and `beta` are roots of quadratic equation `3x^2+8x+2=0`, then find the value of `alpha^3+beta^3`
Solution:
`3x^2+8x+2=0`
Comparing the given equation with `ax^2+bx+c=0`
We get `a=3,b=8,c=2`
Sum of roots `=alpha+beta=(-b)/a=(-8)/3`
Product of roots `=alpha*beta=c/a=2/3`
Now we have to find `alpha^3+beta^3`
`alpha^3+beta^3=(-368)/27`
We know that
`alpha^3+beta^3=(alpha+beta)^3-3alphabeta(alpha+beta)`
`:.alpha^3+beta^3=((-8)/3)^3-3*2/3*(-8)/3`
`:.alpha^3+beta^3=(-512)/27+16/3`
`:.alpha^3+beta^3=(-368)/27`
`:.alpha^3+beta^3=(-368)/27`
4. If `alpha` and `beta` are roots of quadratic equation `3x^2+8x+2=0`, then find the value of `alpha/beta^2+beta/alpha^2`
Solution:
`3x^2+8x+2=0`
Comparing the given equation with `ax^2+bx+c=0`
We get `a=3,b=8,c=2`
Sum of roots `=alpha+beta=(-b)/a=(-8)/3`
Product of roots `=alpha*beta=c/a=2/3`
Now we have to find `alpha/beta^2+beta/alpha^2`
`=(alpha^3+beta^3)/(beta^2alpha^2)`
`alpha^3+beta^3=(-368)/27`
We know that
`alpha^3+beta^3=(alpha+beta)^3-3alphabeta(alpha+beta)`
`:.alpha^3+beta^3=((-8)/3)^3-3*2/3*(-8)/3`
`:.alpha^3+beta^3=(-512)/27+16/3`
`:.alpha^3+beta^3=(-368)/27`
`beta^2alpha^2=4/9`
`:.(alpha^3+beta^3)/(beta^2alpha^2)=((-368)/27)/(4/9)=(-92)/3`
`:.alpha/beta^2+beta/alpha^2=(-92)/3`
5. If `alpha` and `beta` are roots of quadratic equation `3x^2+8x+2=0`, then find the value of `alphabeta^2+betaalpha^2`
Solution:
`3x^2+8x+2=0`
Comparing the given equation with `ax^2+bx+c=0`
We get `a=3,b=8,c=2`
Sum of roots `=alpha+beta=(-b)/a=(-8)/3`
Product of roots `=alpha*beta=c/a=2/3`
Now we have to find `alphabeta^2+betaalpha^2`
`=alphabeta(beta+alpha)`
`:.alphabeta(beta+alpha)=2/3xx(-8)/3=(-16)/9`
`:.alphabeta^2+betaalpha^2=(-16)/9`
This material is intended as a summary. Use your textbook for detail explanation.
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