Question #ec0d2

3 Answers
Apr 1, 2017

Answer:

See the Proof given in the Explanation Section below.

Explanation:

Let the Roots of the given Quadr. be #alpha and beta.#

By what has been given, we may, assume that, #beta=alpha^2.#

It is also known that, #alpha+beta=-b/a, and, alpha*beta=c/a.#

#beta=alpha^2, &, alpha*beta=c/a :. alpha^3=c/a :. alpha=(c/a)^(1/3).#

Hence, #beta=alpha^2=(c/a)^(2/3).#

With #alpha=(c/a)^(1/3), beta=(c/a)^(2/3), and, alpha+beta=-b/a,# we have,

#c^(1/3)/a^(1/3)+c^(2/3)/a^(2/3)=-b/a#

#rArr {c^(1/3)a^(1/3)+c^(2/3)}/a^(2/3)=-b/a.#

#rArr c^(1/3)(a^(1/3)+c^(1/3))=(-b/a)a^(2/3)=-b/a^(1/3).#

#rArr a^(1/3)+c^(1/3)=-b/(a^(1/3)c^(1/3)).#

#rArr (a^(1/3)+c^(1/3))^3={-b/(a^(1/3)c^(1/3))}^3......(ast)#

#rArr (a^(1/3))^3+(c^(1/3))^3+3*a^(1/3)*c^(1/3)(a^(1/3)+c^(1/3))=-b^3/(ac).#

#:. a+c+3*a^(1/3)*c^(1/3){-b/(a^(1/3)c^(1/3))}=-b^3/(ac)...[because,(ast)].#

#rArr a+c-3b=-b^3/(ac).#

# rArr ac(a+c)-3abc=-b^3, or, b^3+ac(a+c)=3abc.#

This completes the Proof.

Enjoy Maths.!

Apr 1, 2017

Answer:

See below.

Explanation:

if #ax^2+bx+c=0# then

#a=-(bx+c)/x^2#
#b=-(ax^2+c)/x#
#c=-(a x^2+bx)#

substituting into

#b^3 + a c (a + c) - 3 a b c=0#

we obtain the polynomial

#p(x)=(ax^2+ax+b) (a^2x^2+(2ab-a^2)x+b^2)#

and for #p(x)=0#

#{(ax^2+ax+b=0),(a^2x^2+(2ab-a^2)x+b^2=0):}#

with the respective roots

#{((-a pm sqrt[a] sqrt[a - 4 b])/(2 a)),((a^2 pm a^(3/2) sqrt[a - 4 b] - 2 a b)/(2 a^2)):}#

and as we can verify

#((-a + sqrt[a] sqrt[a - 4 b])/(2 a))^2 = ( a^2 - a^(3/2) sqrt[a - 4 b] - 2 a b)/(2 a^2)#

or

#((-a - sqrt[a] sqrt[a - 4 b])/(2 a))^2 = ( a^2 + a^(3/2) sqrt[a - 4 b] - 2 a b)/(2 a^2)#

concluding:

The assertion is true. If #b^3 + a c (a + c) - 3 a b c=0# then
concerning the roots of

#ax^2+bx+c=0#, one is the square of the other.

Apr 1, 2017

Suppose the roots of the general quadratic equation:

# ax^2+bx+c = 0 #

are #alpha# and #beta# , then using the root properties we have:

# "sum of roots" \ \ \ \ \ \= alpha+beta = -b/a #
# "product of roots" = alpha beta \ \ \ \ = c /a #

If one root is the square of the other than we can denote the roots by #alpha# and #beta=alpha^2#, and so:

Using the sum of the roots property:

# alpha + beta = -b/a => alpha + alpha^2 = -b/a#
# :. b = -a alpha(1 + alpha) .... [1]#

And the product of the rots property:

# alpha beta = c/a => alpha*alpha^2 = c/a => alpha^3 = c/a#
# :. c = a alpha^3 .... [2]#

Now, consider the LHS of the given identity, and substitute for #b# and #c# using [1] and [2] respectively:

# LHS = b^3+ac(a+c) #
# " " = (-a alpha(1 + alpha))^3 + a(a alpha^3)(a+a alpha^3)#
# " " = -a^3 alpha^3(1 + alpha)^3 + a(a alpha^3)(a)(1 +alpha^3)#
# " " = -a^3 alpha^3(1 + alpha)^3 + a^3 alpha^3(1 +alpha^3)#
# " " = -a^3 alpha^3{(1 + alpha)^3-(1 +alpha^3)}#
# " " = -a^3 alpha^3{(1 +3 alpha + 3 alpha^2 + alpha^3)-(1 +alpha^3 )}#
# " " = -a^3 alpha^3(3 alpha + 3 alpha^2)#
# " " = -3 a^3 alpha^3(alpha + alpha^2)#

But we showed earlier that #alpha^3 = c/a# and #alpha + alpha^2 = -b/a# #

Therefore.

# LHS = -3a^3(c/a)(-b/a)#
# " " = (3a^3bc)/a^2#
# " " = 3abc \ \ \ # QED