Proving Simultaneous Equations Always Have Two Solutions | Formula Included

[jinx007]

Show that the simultaneous equation always have two distinct solutions, for all possible values of k

2x^2+ xy = 10

x + y = k

I know that i have to use b^2 - 4ac = 0

please try the number i am having some problems...
my work

y = k - x

so i replace in equation and i got:

2x^2 + xk - x^2 - 10 = 0 as from here i am stuck..help how to prove that..?

 

[thrill3rnit3]

2x² + kx - x² - 10 = 0

simplifies to

x² - kx - 10 = 0

which is a quadratic equation for any k

the quadratic formula has the +/- thing, so you either have 2 real roots or 2 complex roots [ depending on the value of the discriminant ]

 

[jinx007]

2x² + kx - x² - 10 = 0

simplifies to

x² - kx - 10 = 0

which is a quadratic equation for any k

the quadratic formula has the +/- thing, so you either have 2 real roots or 2 complex roots [ depending on the value of the discriminant ]

hey how do you obtain -kx

 

[thrill3rnit3]

I meant +kx sorry...

But the rest is still the same.

 

[jinx007]

I meant +kx sorry...

But the rest is still the same.

Yeahh me too i have reacheed this level...but the problem is that when i used the formula b^2 - 4ac i cannot get the answer..!


a = -1

b = -k

c = 10

(-k)^2 - 4 (-1)(10) = 0

k^2 = -40

so i am stuck here...what should i do

 

[Borek]

You don't solve for k, you solve for x,y - and you don't give answer as a number (you can't not knowing value of k parameter), but as a formula containing k.

 

[HallsofIvy]

Yeahh me too i have reacheed this level...but the problem is that when i used the formula b^2 - 4ac i cannot get the answer..!


a = -1

b = -k

c = 10

(-k)^2 - 4 (-1)(10) = 0

k^2 = -40

so i am stuck here...what should i do

$b^2- 4ac= k^2+ 40$ which is always positive.

 

[praharmitra]

You do not have to solve for any value. All you have to do is show that there are two distinct solutions for any value of k. This means you must show that discriminant

$$D = b^2 - 4 a c = k^2 + 40 > 0$$ for all values of k. Is this true?