Question

using the quadratic formula,solve the equation a^2 b^2 x^2 -(4b^4 - 3a^4)x - 12a^2 b^2= 0

Answer 1

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Answer 2

GIVEN:-

$\bf \: •Equation \: a²b²x²-(4b²-3a²)x-12a²b²$

TO FIND OUT:-

$\rm \bf \: • \: Roots \: of \: the \: given \: equation=?$

SOLUTION:-

$\rm \bf \: comparing \: the \: given \: equation \: with \: Ax²+Bx+C=0,we \: get$

$\bf \: A=a {}^{2} b {}^{2} ,B=-(4b⁴-3a)=3a²-4b²$

$\bf \: And \: C=-12a²b²$

$\bf \: D=(B²-4AC)$

$\bf \:=(3a⁴-4b⁴)²-4a²b²(-12a²b²)</u><u>\</u><u>\</u><u>$

$\bf = 9a⁸+16b⁸ - 24a⁴b⁴+48a⁴b⁴$

$\bf \: = 9a {}^{2} + 16b {}^{2} + 24a {}^{4} b {}^{4}$

$\bf = (3a {}^{4} ) {}^{2} +( 4b {}^{4} ) {}^{2} 2 \times 3a {}^{4} \times4b {}^{4}$

$\boxed{ \bf = (3a {}^{4} + 4b {}^{4} ) \geqslant 0}$

$\bf \: Hence \: ,the \: given \: equation \: has \: real \: roots \: ,given \: by</u><u>\</u><u>\</u><u>$

$\bf \alpha = </u><u>\</u><u>large</u><u> \frac{ - B + \sqrt{D} }{2A}</u><u>$

$\bf \: = \large\frac{(4b⁴-3a²)+(3a⁴+4b⁴)} {2a²b²}</u><u>$

$\bf= \large \frac{8b {}^{4} }{2a {}^{2}b {}^{2} } = \frac{4b {}^{2} }{a {}^{2} }$

$\bf\:a nd \: \beta= \large \frac{-B- \sqrt{D} }{2A}$

$\bf = \large \frac{(4b {}^{4} - 3a {}^{4} ) - (3a {}^{4} + 4b {}^{4} )}{2a {}^{2}b {}^{2} }$

$\bf \: = \large \frac{ - 6a {}^{4} }{2a {}^{2}b {}^{2} } = \frac{ - 3a {}^{2} }{b {}^{2} }$

$\bf \: Hence \: roots \: of \: the \: given \: equation \: are$

$\boxed{ \bf \: \frac{4b {}^{2} }{a {}^{2} } and \: \frac{ - 3a {}^{2} }{b {}^{2} } }$