Question

Solve the following equation :
a2b2x2 - (4b4 - 3a4)x – 12a2b2 = 0​

Answer 1

Solution: To factorise :- a²b²x² - (4b⁴ - 3a⁴)x - 12a²b²

→ a²b²x² - 4b⁴x + 3a⁴x - 12a²b² = 0

→ b²x(a²x - 4b²) + 3a²(a²x - 4b²) = 0

→ (b²x + 3a²)(a²x - 4b²) = 0

→ (b²x + 3a²) = 0 and (a²x - 4b²) = 0

→ x = - 3a²/b² and x = 4b²/a²

Hence value of x is - 3a²/b & 4b²/a².

Answer 2

Answer:

${\boxed{\bold\green{x =\frac{ -3{a}^{2} }{{b}^{2}}}}}$

Or,

${\boxed{\bold\green{x =\frac{4{b}^{2} }{{a}^{2} } }}}$

Step-by-step explanation:

Question:-

Solve the following equation:-

$\tt {a}^{2} {b}^{2} {x}^{2} -(4{b}^{4} - 3{a}^{4}) x - 12{a}^{2} {b}^{2} = 0$

$\rightarrow\tt {a}^{2} {b}^{2} {x}^{2} - 4{b}^{4}x + 3{a}^{4}x - 12{a}^{2} {b}^{2} = 0 \\\\\rightarrow\tt {b}^{2}x ({a}^{2} x - 4{b}^{2} ) + 3{a}^{2} ({a}^{2}x - 4{b}^{2}) = 0 \\\\\rightarrow\tt ({b}^{2}x + 3{a}^{2} ) ({a}^{2}x - 4{b}^{2}) = 0 \\\\\rightarrow\tt ({b}^{2}x +3{a}^{2} ) = 0 \: \: \: or \: \: \: ({a}^{2}x - 4{b}^{2}) = 0 \\\\\rightarrow\tt {b}^{2}x = -3{a}^{2} \: \: or \: \: {a}^{2}x = 4{b}^{2} \\\\\rightarrow\tt x =\frac{-3{a}^{2}} {{b}^{2} } \: \: \: or \: \: \: x =\frac{4{b}^{2}}{{a}^{2}}$

$\therefore\bold{X = \frac{-3{a}^{2}}{{b}^{2}} \: \: or, {X =\frac{4{b}^{2}}{{a}^{2}} } }$