Question ec0d2

Apr 1, 2017

See the Proof given in the Explanation Section below.

Explanation:

Let the Roots of the given Quadr. be $\alpha \mathmr{and} \beta .$

By what has been given, we may, assume that, $\beta = {\alpha}^{2.}$

It is also known that, $\alpha + \beta = - \frac{b}{a} , \mathmr{and} , \alpha \cdot \beta = \frac{c}{a} .$

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

Hence, $\beta = {\alpha}^{2} = {\left(\frac{c}{a}\right)}^{\frac{2}{3}} .$

With $\alpha = {\left(\frac{c}{a}\right)}^{\frac{1}{3}} , \beta = {\left(\frac{c}{a}\right)}^{\frac{2}{3}} , \mathmr{and} , \alpha + \beta = - \frac{b}{a} ,$ we have,

${c}^{\frac{1}{3}} / {a}^{\frac{1}{3}} + {c}^{\frac{2}{3}} / {a}^{\frac{2}{3}} = - \frac{b}{a}$

$\Rightarrow \frac{{c}^{\frac{1}{3}} {a}^{\frac{1}{3}} + {c}^{\frac{2}{3}}}{a} ^ \left(\frac{2}{3}\right) = - \frac{b}{a} .$

$\Rightarrow {c}^{\frac{1}{3}} \left({a}^{\frac{1}{3}} + {c}^{\frac{1}{3}}\right) = \left(- \frac{b}{a}\right) {a}^{\frac{2}{3}} = - \frac{b}{a} ^ \left(\frac{1}{3}\right) .$

$\Rightarrow {a}^{\frac{1}{3}} + {c}^{\frac{1}{3}} = - \frac{b}{{a}^{\frac{1}{3}} {c}^{\frac{1}{3}}} .$

$\Rightarrow {\left({a}^{\frac{1}{3}} + {c}^{\frac{1}{3}}\right)}^{3} = {\left\{- \frac{b}{{a}^{\frac{1}{3}} {c}^{\frac{1}{3}}}\right\}}^{3.} \ldots . . \left(\ast\right)$

$\Rightarrow {\left({a}^{\frac{1}{3}}\right)}^{3} + {\left({c}^{\frac{1}{3}}\right)}^{3} + 3 \cdot {a}^{\frac{1}{3}} \cdot {c}^{\frac{1}{3}} \left({a}^{\frac{1}{3}} + {c}^{\frac{1}{3}}\right) = - {b}^{3} / \left(a c\right) .$

$\therefore a + c + 3 \cdot {a}^{\frac{1}{3}} \cdot {c}^{\frac{1}{3}} \left\{- \frac{b}{{a}^{\frac{1}{3}} {c}^{\frac{1}{3}}}\right\} = - {b}^{3} / \left(a c\right) \ldots \left[\because , \left(\ast\right)\right] .$

$\Rightarrow a + c - 3 b = - {b}^{3} / \left(a c\right) .$

$\Rightarrow a c \left(a + c\right) - 3 a b c = - {b}^{3} , \mathmr{and} , {b}^{3} + a c \left(a + c\right) = 3 a b c .$

This completes the Proof.

Enjoy Maths.!

Apr 1, 2017

See below.

Explanation:

if $a {x}^{2} + b x + c = 0$ then

$a = - \frac{b x + c}{x} ^ 2$
$b = - \frac{a {x}^{2} + c}{x}$
$c = - \left(a {x}^{2} + b x\right)$

substituting into

${b}^{3} + a c \left(a + c\right) - 3 a b c = 0$

we obtain the polynomial

$p \left(x\right) = \left(a {x}^{2} + a x + b\right) \left({a}^{2} {x}^{2} + \left(2 a b - {a}^{2}\right) x + {b}^{2}\right)$

and for $p \left(x\right) = 0$

$\left\{\begin{matrix}a {x}^{2} + a x + b = 0 \\ {a}^{2} {x}^{2} + \left(2 a b - {a}^{2}\right) x + {b}^{2} = 0\end{matrix}\right.$

with the respective roots

$\left\{\begin{matrix}\frac{- a \pm \sqrt{a} \sqrt{a - 4 b}}{2 a} \\ \frac{{a}^{2} \pm {a}^{\frac{3}{2}} \sqrt{a - 4 b} - 2 a b}{2 {a}^{2}}\end{matrix}\right.$

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 ${b}^{3} + a c \left(a + c\right) - 3 a b c = 0$ then
concerning the roots of

$a {x}^{2} + b x + c = 0$, one is the square of the other.

Apr 1, 2017

Suppose the roots of the general quadratic equation:

$a {x}^{2} + b x + c = 0$

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

$\text{sum of roots} \setminus \setminus \setminus \setminus \setminus \setminus = \alpha + \beta = - \frac{b}{a}$
$\text{product of roots} = \alpha \beta \setminus \setminus \setminus \setminus = \frac{c}{a}$

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:

$\alpha + \beta = - \frac{b}{a} \implies \alpha + {\alpha}^{2} = - \frac{b}{a}$
$\therefore b = - a \alpha \left(1 + \alpha\right) \ldots . \left[1\right]$

And the product of the rots property:

$\alpha \beta = \frac{c}{a} \implies \alpha \cdot {\alpha}^{2} = \frac{c}{a} \implies {\alpha}^{3} = \frac{c}{a}$
$\therefore c = a {\alpha}^{3} \ldots . \left[2\right]$

Now, consider the LHS of the given identity, and substitute for $b$ and $c$ using  and  respectively:

$L H S = {b}^{3} + a c \left(a + c\right)$
$\text{ } = {\left(- a \alpha \left(1 + \alpha\right)\right)}^{3} + a \left(a {\alpha}^{3}\right) \left(a + a {\alpha}^{3}\right)$
$\text{ } = - {a}^{3} {\alpha}^{3} {\left(1 + \alpha\right)}^{3} + a \left(a {\alpha}^{3}\right) \left(a\right) \left(1 + {\alpha}^{3}\right)$
$\text{ } = - {a}^{3} {\alpha}^{3} {\left(1 + \alpha\right)}^{3} + {a}^{3} {\alpha}^{3} \left(1 + {\alpha}^{3}\right)$
$\text{ } = - {a}^{3} {\alpha}^{3} \left\{{\left(1 + \alpha\right)}^{3} - \left(1 + {\alpha}^{3}\right)\right\}$
$\text{ } = - {a}^{3} {\alpha}^{3} \left\{\left(1 + 3 \alpha + 3 {\alpha}^{2} + {\alpha}^{3}\right) - \left(1 + {\alpha}^{3}\right)\right\}$
$\text{ } = - {a}^{3} {\alpha}^{3} \left(3 \alpha + 3 {\alpha}^{2}\right)$
$\text{ } = - 3 {a}^{3} {\alpha}^{3} \left(\alpha + {\alpha}^{2}\right)$

But we showed earlier that ${\alpha}^{3} = \frac{c}{a}$ and $\alpha + {\alpha}^{2} = - \frac{b}{a}$

Therefore.

$L H S = - 3 {a}^{3} \left(\frac{c}{a}\right) \left(- \frac{b}{a}\right)$
$\text{ } = \frac{3 {a}^{3} b c}{a} ^ 2$
$\text{ } = 3 a b c \setminus \setminus \setminus$ QED