Question 97013

Mar 7, 2016

The solution is $x = - 0.375$

Explanation:

[−8(3m^2−2m)]^2−4(1+4m^2)(36m^2−48m−9)=0

[−8 color(blue)((3m^2−2m))]^2 =4(1+4m^2)(36m^2−48m−9)

Note:
As $\textcolor{b l u e}{\left(m\right)}$ is common to both the terms of the L.H.S ($3 {m}^{2}$ and $2 m$ )we take it out of the bracket.

[−8 color(blue)(m) (3m−2)]^2 =4(1+4m^2) color(purple)((36m^2−48m−9)

Note:
color(purple)((3) is common to all the terms within the second bracket of the
R.H.S ($36 {m}^{2}$, −48m and −9 ) so, we take it out of the bracket.

[−8 m (3m−2)]^2 =4(1+4m^2) * color(purple)((3)) (12m^2−16m−3)

(−8 m)^2 (3m−2)^2 =4 xx color(purple)((3)) (1+4m^2) * (12m^2−16m−3)

cancel64m^2 color(green)((3m−2))^2 =cancel12 (1+4m^2) *(12m^2−16m−3)

16m^2 color(green)((3m−2))^2 =3 (1+4m^2) * (12m^2−16m−3)

Applying the property: color(green)((a-b)^2= a^2-2ab +b^2 to color()((3m−2))^2

(16m^2) color(green)((9m^2-12m +4)) =3 (1+4m^2) * (12m^2−16m−3)

(16m^2) * (9m^2) + (16m^2) * ( -12m) +(16m^2) * (4) =3 (1+4m^2) xx (12m^2−16m−3)

144m^4 -192m^3 +64m^2 =3 (color(blue)(1)+color(purple)(4m^2)) xx (12m^2−16m−3)

144m^4 -192m^3 +64m^2 =3 [(color(blue)(1 * (12m^2) + 1 * (-16m) + 1 * (-3)) + color(purple)(4m^2 *(12m^2) + 4m^2 (-16m) + 4m^2 (-3)]

144m^4 -192m^3 +64m^2 =3 [(12m^2 -16m -3 +48m^4 -64m^3 -12m^2]

144m^4 -192m^3 +64m^2 =3 [(cancel12m^2 -16m -3 +48m^4 -64m^3 -cancel12m^2]

$144 {m}^{4} - 192 {m}^{3} + 64 {m}^{2} = 3 \left(- 16 m - 3 + 48 {m}^{4} - 64 {m}^{3}\right)$

$144 {m}^{4} - 192 {m}^{3} + 64 {m}^{2} = 3 \cdot \left(- 16 m\right) + 3 \cdot \left(- 3\right) + 3 \cdot \left(48 {m}^{4}\right) + 3 \cdot \left(- 64 {m}^{3}\right)$

$144 {m}^{4} - 192 {m}^{3} + 64 {m}^{2} = - 48 m - 9 + 144 {m}^{4} - 192 {m}^{3}$

$\cancel{144} {m}^{4} - \cancel{192} {m}^{3} + 64 {m}^{2} = - 48 m - 9 + \cancel{144} {m}^{4} - \cancel{192} {m}^{3}$

$64 {m}^{2} = - 48 m - 9$

$64 {m}^{2} + 48 m + 9 = 0$

The equation is of the form color(blue)(am^2+bm+c=0# where:
$a = 64 , b = 48 , c = 9$

The Discriminant is given by:

$\Delta = {b}^{2} - 4 \cdot a \cdot c$

$= {\left(48\right)}^{2} - \left(4 \cdot 64 \cdot 9\right)$

$= 2304 - 2304 = 0$

The solution is found using the formula
$x = \frac{- b \pm \sqrt{\Delta}}{2 \cdot a}$

$x = \frac{\left(- 48\right) \pm \sqrt{0}}{2 \cdot 64} = \frac{- 48 \pm 0}{128}$

$x = - \frac{48}{128}$

$x = - 0.375$