Question

solve for x :
$9 {x}^{2} - 6 {a}^{2} x + ( {a}^{4} - {b}^{4} ) = 0$

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Answer 1

$\huge\tt\red{\bold{\underline{\underline{❥Question᎓}}}}$solve for x :

$9 {x}^{2} - 6 {a}^{2} x + ( {a}^{4} - {b}^{4} ) = 0$

$\huge\mathcal{Answer}$

$\bold{ GIVEN}$:$9 {x}^{2} - 6 {a}^{2} x + ( {a}^{4} - {b}^{4} ) = 0$

$⟹</p><p>9 {x}^{2} - 6 {a}^{2} x + ( {a}^{4} - {b}^{4} ) = 0$

$⟹</p><p>9 {x}^{2} - 6 {a}^{2} x + ( {a}^{2} + {b}^{2} )( {a}^{2} - {b}^{2} ) = 0$

$⟹</p><p>\bold{Here\:product \:of \:coefficient \:of \:x \:square \:and \:the \:constant \:term}$

$= 9( {a}^{2} + {b}^{2} )( {a}^{2} - {b}^{2} )$

$= (3 {a}^{2} + 3 {b}^{2} )(3 {a}^{2} - 3 {b}^{2} )$

$so \: that \: 3 {a}^{2} + 3 {b}^{2} + 3 {a}^{2} - 3 {b}^{2} = 6 {a}^{2}$

$⟹</p><p>\bold{∴ The\: given\: quadratic \:equation\: can\: be\: written\: as }$

$⟹</p><p>9 {x}^{2} - (3( {a}^{2} + {b}^{2} ) + 3( {a}^{2} - {b}^{2} ))x + ( {a}^{2} + {b}^{2} )( {a}^{2} - {b}^{2} ) = 0$

$⟹</p><p>9 {x}^{2} - 3x( {a}^{2} + {b}^{2} ) - 3x( {a}^{2} - {b}^{2} ) + ( {a}^{2} + {b}^{2} )( {a}^{2} - {b}^{2} ) = 0$

$⟹</p><p>3x(3x - {a}^{2} - {b}^{2} ) - ( {a}^{2} - {b}^{2} )(3x - {a}^{2} - {b}^{2} ) = 0$

$⟹</p><p>( 3x - {a}^{2} - {b}^{2} )(3x - {a}^{2} + {b}^{2} ) = 0$

$⟹</p><p>3x - {a}^{2} - {b}^{2} = 0$

$⟹</p><p>3x - {a}^{2} + {b}^{2} = 0$

$⟹</p><p>3x = {a}^{2} + {b}^{2}$

$⟹</p><p>3x = {a}^{2} - {b}^{2}$

$⟹</p><p>x = \frac{ {a}^{2} + {b}^{2} }{3}$

$⟹</p><p> = \frac{ {a}^{2} - {b}^{2} }{3}$

$\bold{∴The\: solution\: of \:given\: quadratic \:equation\: are:-}$

$\bold{\red{ \frac{ {a}^{2} + {b}^{2} }{3} \: and \: \frac{ {a}^{2} - {b}^{2} }{3} }}$

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Answer 2

Answer:

Answer:

9x^-6b^2x-(a^4-b^4)

9x^2-6b^2x-a^4+b^4

By quadratic formula:

D=6b^2)^2-4×9×(-a^4+b^4)

D=36b^4+36a^4-36b^4

D=36a^4

Now,

X=-(-6b^2)+-√36a^4/2×9

X=6b^2+-6a^2/18

X=6b^2+6a^2/18

X=6(b^2+a^2)/18

X=b^2+a^2/3

X=6b^2-6a^2/18

X=6(b^2-a^2)/18

X=b^2-a^2/3

X=-(a^2-b^2)/3

So, the value of X=a^2+b^2/3,

X=-(a^2-b^2)/3

Hope it's help you