Questions of Quadratic Equations and thier Roots

[Mr.maniac]

Homework Statement

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

Homework Equations

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

 

[Mr.maniac]

after that i got stuck...

 

[Mr.maniac]

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

 

[Mark44]

For some reason, this thread was marked "Solved," but it doesn't appear to me that it actually is solved.

Homework Statement

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

What do you mean by "roots of x²"? The equation $x^2 = 0$ has only 0 as a root.

3)If x²-ax-21=0 and x²-3ax+35 have a root in commom

Homework Equations

The Attempt at a Solution

I took the roots as p q r s

?
A quadratic equation has at most two real roots. What do your variables p, q, r, and s represent?

by the relations between co efficients and roots i got
1) p+q= -k/2
pq= -5/2
p+r=3
pr= -4

The question asks for k.
I solved this one by factoring the second equation, $x^2 - 3x - 4 = 0$.

2)similar to q1
p+q=3
pq=2m
2p+r=1
2pr=m

I don't understand the question. Possibly you have a typo in it.

3)again similar to q1
p+q=a
pq=-21
p+r=3a
pr=35

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?

 

[epenguin]

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

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
2α² +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. )