Solve the given equation that involves fractional exponent

[chwala]

Homework Statement

$5(x+1)^{1.5}-4x-28=0$

Relevant Equations

Algebra.

Is there a better way of solving this,
My steps,
$5(x+1)^{1.5}-4x-28=0$
$(x+1)^{1.5}=\dfrac{4x+28}{5}$
$(x+1)^3=(\dfrac{4x+28}{5})^2$
$x^3+3x^2+3x+1=\dfrac{16(x+7)^2}{25}$
$25x^3+75x^2+75x+25=16(x^2+14x+49)$
$25x^3+59x^2-149x-759=0$

Letting
$f(x)=25x^3+59x^2-149x-759=0$

and using factor theorem,

$f(3)=0$

$⇒x=3$.

 

[fresh_42]

Homework Statement: $5(x+1)^{1.5}-4x-28=0$
Relevant Equations: Algebra.

Is there a better way of solving this,
My steps,
$5(x+1)^{1.5}-4x-28=0$
$(x+1)^{1.5}=\dfrac{4x+28}{5}$
$(x+1)^3=(\dfrac{4x+28}{5})^2$
$x^3+3x^2+3x+1=\dfrac{16(x+7)^2}{25}$
$25x^3+75x^2+75x+25=16(x^2+14x+49)$
$25x^3+59x^2-149x-759=0$

Letting
$f(x)=25x^3+59x^2-149x-759=0$

and using factor theorem,

$f(3)=0$

$⇒x=3$.

How did you find $x=3$ and what about the two other roots? You also made an irreversible step by squaring. Are the two other roots solutions of the original equation, too?

 

[chwala]

How did you find $x=3$ and what about the two other roots? You also made an irreversible step by squaring. Are the two other roots solutions of the original equation, too?

...by trial and error method around $x=±0, ±1, ±2, ...$. The other roots are imaginary.

By trial and error, i had already checked using Newton's method and could see where the approximation value was headed to. You may advise on a better way that will incorporate the imaginary solution/s.

 

[fresh_42]

The other roots are imaginary.

Yes, but they could still provide solutions. I don't think so, but it cannot be ruled out. You haven't said that only real solutions count.

 

[fresh_42]

I checked the complex solutions
$$
x\in \dfrac{1}{25}\cdot\left\{-67\mp 6\sqrt{51}\;\mathrm{i}\right\}
$$
and it was close:
$$
\dfrac{4}{5}\cdot \dfrac{x+7}{x+1}\in\dfrac{1}{5}\cdot\left\{-3\pm \sqrt{51}\;\mathrm{i}\right\} \quad\text{ and }\quad \sqrt{x+1}\in \dfrac{1}{5}\cdot \left\{3 \mp \sqrt{51} \;\mathrm{i}\right\}
$$
I hope I made no sign error. $-\sqrt{x+1}=\dfrac{4}{5}\cdot \dfrac{x+7}{x+1}$ would have been the other solutions! This means that we made the (silent) assumption that $(x+1)^{0.5}=+\sqrt{x+1}.$