Quadratic equation Problem (2 Viewers)

[movie_detective]

Hey all,

The problem is to : Form the quadratic equation with roots which exceed by 2 the roots of *3x^2-(p-4)^2x -(2p-1)= 0. Also find the values of p for which the given equation has equal roots.

Thx all

 

[asianese]

The easiest thing here to do is to use the relationship between the roots and the coefficients.

Say the roots of the given equation are A and B. We wish to form a quadratic with roots (A+2) and (B+2). Any regular quadratic equation has the form: y^2 - (sum of roots)y + (product of roots)=0. So the equation is y^2 - ((A+2)+(B+2))y+(A+2)(B+2)=0.

So insert the values of A+B and AB from the given equation into the new equation.


For equal roots, the discriminant is 0!

 

[karnbmx]

1st part:

consider roots of given quadratic as alpha, beta:
therefore, alpha + beta = ((p-4)^2)/3 (sum of roots one at a time)
also alpha*beta = (1-2p)/3 (product of roots)

NOW let the roots of quadratic we are looking for be alpha + 2, beta + 2
thus sum of roots yields alpha + beta + 4 = -b/a

BUT we know that alpha + beta = ((p-4)^2)/3

so -b/a = ((p-4)^2)/3 + 4 thus b/a = -((p-4)^2)/3 - 4 which is the second term of the quadratic

similarly look for the third term and you get your answer.

Part 2:

This part is easier. Consider the fact that because all the roots are EQUAL, the discriminant must equal to zero.

Therefore, from there: b^2 - 4ac = 0

solve from there using coefficients from the quadratic given.

 

[karnbmx]

The easiest thing here to do is to use the relationship between the roots and the coefficients.

Say the roots of the given equation are A and B. We wish to form a quadratic with roots (A+2) and (B+2). Any regular quadratic equation has the form: y^2 - (sum of roots)y + (product of roots)=0. So the equation is y^2 - ((A+2)+(B+2))y+(A+2)(B+2)=0.

So insert the values of A+B and AB from the given equation into the new equation.


For equal roots, the discriminant is 0!

DAMN you beat me to it.

 

[asianese]

.

I was gonna latex, but I knew I'd be beaten.

 

[movie_detective]

Thx for both replying. Just for part b, I'm not sure if this is correct however the equation seems unsolvable therefore the given equation doesn't have equal roots ?

 

[asianese]

Check your equation again.

Did you mean: ?

 

[movie_detective]

Yeah, that's the equation however b^2-4ac=0 seems unsolvable

 

[asianese]

Yeah it's unsolvable ...