Question

if the equation x^2-cx+d-0 has roots equal to the fourth power of the roots of x^2+ax+b=0 where a^2>4b then the roots of x^2-4bx+2b^2-c=0 will be
a.both real
b.both negative
c.both positive
d.one positive and one negative

Answer 1

@hartnn pls..hlp

Answer 2

let the roots of x^2-cx+d=0 is α and β while the roots of x^2+ax+b=0 is p and q. according information above, known α = p^4 and β = q^4.

now from the equation of x^2-cx+d=0 we get α + β = c while from the equation of x^2+ax+b=0 we get p+q = c and pq = b

because α = p^4 and β = q^4.
so, α + β = p^4 + q^4. simplied
α + β = (p^2 + q^2)^2 - 2(pq)^2
α + β = ((p + q)^2 - 2pq)^2 - 2(pq)^2

distribute the values of α + β = c, p+q=c and pq = b, we get
α + β = ((-a)^2 - 2b)^2 - 2(b)^2
c = a^4 - 4a^2b + 4b^2 - 2b^2
c = a^4 - 4a^2b + 2b^2

now look at the equation x^2-4bx+2b^2-c=0, by complete square it can be ;
x^2-4bx+2b^2-c=0
(x - 2b)^2 - 2b^2 - c = 0
(x - 2b)^2 = 2b^2 + c

again distribute that c = a^4 - 4a^2b + 2b^2, therefore
(x - 2b)^2 = 2b^2 + c
(x - 2b)^2 = 2b^2 +a^4 - 4a^2b + 2b^2
(x - 2b)^2 = a^4 - 4a^2b + 4b^2
complete square again right side :
(x - 2b)^2 = (a^2 - 2b)^2

finally, we get x :
x - 2b = +- sqrt((a^2 - 2b)^2)
x - 2b = +- (a^2 - 2b)
x = 2b +- (a^2 - 2b)

x1 = 2b + a^2 - 2b = a^2
x2 = 2b - a^2 + 2b = 4b - a^2

now, in logically. the value a^2 always be positive. while the value of 4b - a^2 would be negative. (because information above says a^2 > 4b)

Answer 3

so, the answer is D. one positive and one negative

Answer 4

\[ x^2-cx+d=0 \ \ ; \ \ (x-\alpha)(x-\beta)
\\ \ \hspace{125 px} \alpha + \beta = c\]
\[x^2+ax+b=0 \ \ ; \ \ (x+\sigma)(x+\lambda)
\\ \hspace{125px} \sigma \lambda = b
\\ \hspace{125px} \alpha = \sigma^4
\\ \hspace{125px} \beta = \lambda^4\]
\[x^2-4bx+2b^2-c=0
\\ \hspace{125px} D = 16b^2 - 8b^2+4c
\\ \hspace{125px} \ \ = 8b^2 + 4c
\\ \hspace{125px} \ \ = 8 (\sigma \lambda)^2 + 4(\sigma^4 + \lambda^4)\]

\[ \textrm{roots are} \ \ \frac{- \sigma \lambda ±\sqrt{8 (\sigma \lambda)^2 + 4(\sigma^4 + \lambda^4)}}{2}\]
\[ D > -\sigma \lambda\]

So yeah, what RaDen said. It's D Just wanted to finish typing it :P

Answer 5

~~~
roots are
\[2 \sigma \lambda ± \sqrt{2\sigma^2 \lambda^2 + \sigma^4 + \lambda^4}\]

Answer 6

i think that p+q shud be -a hw u wrote p+q=c

Answer 7

sorry, that should p+q = -a not c.
but the next step already be correct