# Questions of Quadratic Equations and thier Roots

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1. Jul 22, 2017

### Mr.maniac

1. The problem statement, all variables and given/known data
1)The value of k, so that the equations 2x2+kx-5=0
and x2-3x-4=0 have one root in common
2)The value of m for which one of the roots of x2 is double of one of roots of x2-x+m=0
3)If x2-ax-21=0 and x2-3ax+35 have a root in commom

2. Relevant equations

3. The attempt at a solution
I took the roots as p q r s
by the relations between co efficients and roots i got
1) p+q= -k/2
pq= -5/2
p+r=3
pr= -4
2)similar to q1
p+q=3
pq=2m
2p+r=1
2pr=m
3)again similar to q1
p+q=a
pq=-21
p+r=3a
pr=35

Last edited: Jul 22, 2017
2. Jul 22, 2017

### Mr.maniac

after that i got stuck....

3. Jul 22, 2017

### Mr.maniac

could someon give me general instructions about these kind of questions?

4. Jul 22, 2017

### Staff: Mentor

For some reason, this thread was marked "Solved," but it doesn't appear to me that it actually is solved.
What do you mean by "roots of x2"? The equation $x^2 = 0$ has only 0 as a root.
???
A quadratic equation has at most two real roots. What do your variables p, q, r, and s represent?
I solved this one by factoring the second equation, $x^2 - 3x - 4 = 0$.
I don't understand the question. Possibly you have a typo in it.
The only possible roots of the first equation are -3 and 7 or 3 and -7. What are the only possible pairs of roots in the second equation?

5. Jul 25, 2017

### epenguin

The general method is just algebraic elimination, of which you have certainly already met other examples in other simultaneous equations.

It is not necessary, but at this stage may be helpful in overcoming the mental blockage* that is preventing you seeing how easy and natural it is to proceed, if you call the particular value of x that makes a couple of equations both true another name - you could call it x1 or you could call it α.

Thus you are being told that there exists a number, α for which both
2 +kα-5=0 and α2-3α-4=0

You can surely from these two get a new equation in which α2 is eliminated and is still true. After which you should be able to go on eliminating until you get an equation in which α does not appear at all.

(There does exist a general expression in their coefficients, called the 'Eliminant' of two polynomials which vanishes when they have a common root or factor, which is 4x4 determinant for two quadratics. But this is just a convenient formulation of the algebraic eliminations you would do anyway. To show things, even things you know, are connected up, the discriminant of a quadratic is just the eliminant of the quadratic and it's derivative.)

(* there is a possible mental blockage of continuing to think of x as 'a Variable', something that could be just anything, Whereas the problem itself has restricted it to something definite even if not yet known. When you're more used to it you can kick away that prop. )

Last edited: Jul 25, 2017