Question

Question #ec0d2

Answer 1

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.!

Answer 2

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.

Answer 3

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