To calculate: The solution set of the polynomial equation 3x2(x2+2)=20−x2.

Question

Drood1980 · Accepted Answer

Given Information:
The provided polynomial equation is $3 x^{2} (x^{2} + 2) = 20 - x^{2}$ .
Formula used:
According to zero product property if,
$m n = 0$ ;
Then,
$m = 0$
Or,
$n = 0$
Calculation:
This is a polynomial equation with one side equal to zero.
$3 x^{2} (x^{2} + 2) = 20 + x^{2} = 0$
Simplify the given equation,
$3 x^{2} (x^{2} + 2) = 20 + x^{2} = 0$
$3 x^{4} + 7 x^{2} - 20 + x^{2} = 0$
$3 x^{4} + 7 x^{2} - 20 = 0$
Let $x^{2} = t$ , the equation becomes,

$3 P + 71 - 20 = 0$

Do the factors of the given equation,
$3 t^{2} - 5 t + 12 t - 20 = 0$
$t (3 t - 5) + 4 (3 t - 5) = 0$
$(3 t - 5) (t - 4) = 0$
Set each factor equal to zero.
$(3 t - 5) = 0$
$t = \frac{5}{3}$
Or
$(t + 4) = 0$
$t = - 4$
Now the value of xis
$x^{2} = t$
Simplify it, $x^{2} = \frac{5}{3} o r x^{2} = - 4$
$x = \pm \sqrt{\frac{5}{3}} o r x = \pm \sqrt{- 4}$
$x = \pm \sqrt{\frac{5}{3}} o r x = \pm 2 i$
Check at $x = \pm \sqrt{\frac{5}{3}}$
$3 {(\pm \sqrt{\frac{5}{3}})}^{2} [{(\pm \sqrt{\frac{5}{3}})}^{2} + 2] 20 - {(\pm \sqrt{\frac{5}{3}})}^{2}$
$5 ∣ \frac{5}{3} + 2 ∣ \begin{matrix} ? \\ = \end{matrix} 20 - \frac{5}{3}$
$\frac{55}{3} = \frac{55}{3}$
The left side value is equal to the right-side value. Thus, the value
$x = \pm \sqrt{\frac{5}{3}}$
Check at $x = \pm 21$
$3 [{(\pm 2 i)}^{2} + 2] 20 - {(\pm 2 i)}^{2}$
$- 12 [- 4 + 2] 20 + 4$
$- 12 [- 2] 24$
$24 = 24$
The left side value is equal to the right-side value. Thus, the value

0

To calculate: The solution set of the polynomial equation 3x^{2}\le

Answered question

Answer & Explanation

New Questions in Pre-Algebra