## Friday, May 31, 2013

Now we will see a few quadratic equations using the formula method.
a) Solve the quadratic equation  5 x 2 - 2 x - 9 = 0.
Solution:
1) Comparing  5 x 2 - 2 x - 9 = 0 with a x 2 + b x + c = 0 we get,
a = 5, b = - 2, c = - 9

b) Solve the quadratic equation  3 x 2 - 4 x - 3 = 0.
Solution:
1) Comparing  3 x 2 - 4 x - 3 = 0  with a x 2 + b x + c = 0 we get,
a = 3, b = - 4, c = - 3

Now we will see some rules about the roots of the quadratic equation.
1) Sum of the roots (α, β) of the quadratic equation a x 2 + b x + c = 0,
α  +  β  =  - b / a  and   α β  =  c / a
α  +  β  =  - Coefficient of x / Coefficient of x 2  and   α β  =  Constant / Coefficient of x 2.
2) If α, β are the roots of the quadratic equation then the quadratic equation will be
x 2 - (Sum of the roots)  x + (Product of the roots) = 0,
x 2 - (α  +  β)  x + (α β) = 0,

Now we will see some important problems on Quadratic equation and the roots of the Quadratic Equations

a) If the sum of the roots of the quadratic equation is 3 and the sum of their cubes is 63, then find the equation.
Solution:
1)  Here, α + β =  3 and  α 3 + β 3 =  63
2)  We know that , α 3 + β 3 =  (α + β)- 3 α β (α + β)
so ,     63      =  (3)- 3 α β (3)
63      =  27 - 9 α β       Dividing by 9 through out we get
7       =  3  - α β
α β   =  3  - 7
α β   =  - 4
3) We know that  the quadratic equation will be
2 - (α  +  β)  x + (α β) = 0
2 - (3)  x + (- 4) = 0
2 - 3 x - 4 = 0
4)  Answer: So the required quadratic equation will be 2 - 3 x - 4 = 0.
Few more problems will be discussed in the next Blog.