Question

parabola y=x^2+2x+L intersects line y=mx+n at points

Answer 1

question continues here!!
parabola y=x^2+2x+L intersects line y=mx+n at points \[\left( \alpha,\gamma1 \right) and \left( \beta,\gamma2 \right). if \alpha + \beta=3 and \alpha \beta=-4 \] then m+n is?

Answer 2

Okay intersection points are found out by solving the equations of both the curves.

So, firs try to solve both the curves.

Answer 3

That is put y = mx +n in equation of parabola.

Answer 4

i cant understand @vishweshshrimali5

Answer 5

Its okay

Answer 6

See:

By solving the equations I mean,

solve this:

mx +n = x^2 + 2x + L

Answer 7

\[\large{x^2 + x(2-m) + (L-n) = 0}\]

Answer 8

Now since the solutions of x are \(\large{\alpha,\beta}\). Find out :

(1) \(\large{\alpha+\beta}\) with the help of above equation.

(2) \(\large{\alpha\beta}\) with the help of above equation.

Answer 9

Can you do this @dinisha ?

Answer 10

is \[\alpha + \beta = 2-m\] and \[\alpha \beta = L-n\]

Answer 11

See:

in \(\large{\alpha + \beta = \cfrac{-(2-m)}{1}}\)

\(\large{\alpha\beta}\) is correct.

Answer 12

Remember that :

sum of roots was -b/a .

Answer 13

so -2 + m ?

Answer 14

Yes

Answer 15

But, you are already given \(\alpha+\beta,\alpha\beta\) in the question. So, with the help of those values find L and m.

Answer 16

how??

Answer 17

See:

in the question \(\large{\alpha+\beta = 3}\)
But, we obtained that:

\(\large{\alpha+\beta = -2 + m}\)

So,

-2+m = 3

=> m = 5

Answer 18

Similarly, find out L

Answer 19

here there are two unknows!! L-n=-4

Answer 20

Yeah sorry

But, you now have one equation also.

Answer 21

one equation??

Answer 22

I meant L-n = -4

Answer 23

Are you sure there was L in the original question ?

Answer 24

I think that's "1" not "L".

Answer 25

Now, the equation (L-n =-4) will become,

1- n = -4
=> n = 5

So, m = 5 and n = 5. Now, can you find m+n ?

Answer 26

Did you get it @dinisha ?

Answer 27

ohh yaa i got it thank you!!