Answer:
[tex]x = \frac{3}{2}orx=-\frac{3}{2}[/tex]
Step-by-step explanation:
The given Equation is
[tex]3x^{2} + 2 = 11-x^{2}[/tex] .................(i)
Adding ( [tex]x^{2} -11[/tex] ) on both sides of (i)
[tex]3x^{2}+2+x^{2}-11 = 11-x^{2}+x^{2}-11[/tex]
solving and cancelling out the terms will give
⇒ [tex]4x^{2} -9=0[/tex]
⇒ [tex](2x)^{2} -(3)^{2} = 0[/tex] ...................(ii)
Now we Know that
[tex](a)^{2} -(b)^{2} = (a-b)(a+b)[/tex]
Applying this on equation (ii)
[tex](2x-3)(2x+3)=0[/tex]
This will lead us to
Either 2x-3 =0 .......(iii) or 2x+3=0 ..........(iv)
Solving equation (iii) for value of x
2x - 3 =0
adding 3 on both sides
2x -3 +3 = 0+ 3
2x = 3
Cross multiplying gives
[tex]x=\frac{3}{2}[/tex]
Solving equation (iv) for value of x
2x + 3 =0
adding -3 on both sides
2x -3 +3 = 0- 3
2x = -3
Cross multiplying gives
[tex]x=\frac{-3}{2}[/tex]
so [tex]x=\frac{3}{2}[/tex] and [tex]x=\frac{-3}{2}[/tex]
