The roots of the equation 2x² +px+q are (2alpha+beta) and (alpha+2Beta).Calculate the value of p & value of q?

Question

please show working.thanks a lot

Answer 1

Consider quadratic equation ax^2 + bx + c = 0

Sum of its roots = -b/a

and product of its roots = c/a

In the given problem, a = 2, b = p, c = q

Sum of roots = -b/a

Or 2alpha+beta + alpha+2beta = -p/2

Or 3alpha + 3beta = -p/2

Or 6alpha + 6beta = -p

Or 6(alpha + beta) = -p

Or p = -6(alpha + beta)

Product of roots = c/a

Or (2alpha+beta)(alpha+2Beta) = q/2

Or q = 2*(2alpha+beta)(alpha+2Beta)

Ans:

p = -6(alpha + beta)

q = 2*(2alpha+beta)(alpha+2Beta)

Answer 2

Simply replace each root in the equation. You replace in turn the x with  (2alpha+beta) and (alpha+2Beta). Thus you have a system with 2 equations and 2 unknowns:

2(2alpha+beta)² + p(2alpha+beta) + q = 0

and                                                                 =>

2(alpha+2beta)² + p(alpha+2beta) + q = 0

p(2alpha+beta) + q = -2(2alpha+beta)²

and                                                                 =>

p(alpha+2beta) + q = -2(alpha+2beta)²  |(-1)

Multiply the second equation with -1 and add the 2 equations term by term. You get the following equation:

p(2alpha+beta) + q - p(alpha+2beta) - q = -2(2alpha+beta)² + 2(alpha+2beta)² =>

p(2alpha + beta - alpha - 2beta) = -2(4alpha² + 4alphabeta + beta²) + 2(alpha² + 4alphabeta + 4beta²)  =>

p(alpha - beta) = -8alpha² -8alphabeta -2beta² + 2alpha² + 8alphabeta + 8beta² =>

p(alpha - beta) = -6alpha² + beta² =>

p(alpha - beta) = -6(alpha² - beta²) =>

p(alpha - beta) = -6(alpha - beta)(alpha + beta) =>

p = -6(alpha + beta)

Thus you found the value for p. Let's consider now the first equation from the system at the beginning of my reply:

2(2alpha+beta)² + p(2alpha+beta) + q = 0

Now we know that p = -6(alpha + beta)

Let's replace p in the equation with this value and find the value for q:

2(2alpha+beta)² + -6(alpha + beta)(2alpha+beta) + q = 0 =>

q = -2(2alpha+beta)² + 6(alpha + beta)(2alpha+beta) =>

q = -2(4alpha² + 4alphabeta + beta²) + 6(2alpha² + alphabeta + 2 alphabeta + beta²) =>

q = -8alpha² - 8alphabeta -2beta² + 12alpha² + 18alphabeta + 6beta² =>

q = 4alpha² + 4beta² + 10alphabeta =>

q = 2(2alpha + beta)(alpha + 2beta)

Thus, the solution is:

p = -6(alpha + beta)

q = 2(2alpha + beta)(alpha + 2beta)

Answer 3

I assume that the "equation" is 2x² +px+q =0 not 2x² +px+q .

Then sum of roots = -p/2 = 3(alpha+beta) so

p = - 6(alpha+beta)

and q/2 = product of roots so q =2[2(aplpha)^2 + 5(alpha)(beta) + 2(beta)^2]

Answer 4

let us call alpha a and beta b. so, the roots are 2a + b and a + 2b.

the sum of the roots is -p/2

(2a+b) + (a+2b) = -p/2

3(a+b) = - p/2

p = - 6(a+b).

the product of the roots is q/2:

(2a+b)*(a+2b) = q/2.

q = 2 * (2a+b) (a+2b).

Answer 5

Create two simultaneous equations.  

1.  Substitute (2 alpha + beta) for x and solve for p.

2.  Substitute (alpha + 2 beta) for x and solve for p.

Solve the simultaneous equations.

Answer 6

just say that the roots are 2a+b and a+2b.

* if we have equation (ax^2+bx+c)=0 and we can factorize it to (dx+e)+(fx+g), we know that the roots are -e/d and -g/f, and we know that a=df, b = ef+dg , c= eg.

-b/a = -e/d -g/f which is first root+ second root,

c/a =eg/df which is first root * second root.

Based on that, now we get :

1. 3a+3b=-p/2

2. (a+2b)(b+2a)=q/2

So, we get the value of p and q:

1.  p=-6a-6b

2. q=2(2a^2+5ab+2b^2)=4a^2+10ab+4b^2