Question

(3x+4)^2+(3x-2)^2=(6x+5)(3x-2)+12
try this one

Answer 1

(3x + 4)² + (3x - 2)² = (6x + 5)(3x - 2 ) + 12

Apply (a ±b)² = a² ± 2ab + b² to the LHS and expand RHS:

9x² + 24x + 16 + 9x² - 12x + 4 = 18x² - 12x + 15x - 10 + 12

Combine like terms:

18x²+ 12x  + 20 = 18x² + 3x + 2

Subtract 18x² from both sides:

12x + 20 = 3x + 2

Subtract 3x from both sides:

9x + 20 = 2

Subtract 20 from both sides:

9x = -18

Divide both sides by 9:

x = -2

Answer: x = -2

Answer 2

Q) (3x+4)^2+(3x-2)^2=(6x+5)(3x-2)+12
=)(3x) ^2+2*3x*4+(4)^2+(3x) ^2-2*3x*2+(2)^2=6x(3x-2)+5(3x-2)+12
=)9x^2+24x+16+9x^2-12x+4=18x^2-12x+15x- 10+12
=)18x^2+12x+20=18x^2+15x-12x+2
=)12x+20=-12x+15x+2
=24x+18=15x
=)9x=18
=)x=2