Solve:-
4x² - 4a²x + (a⁴ - b⁴) = 0
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Solve:-
4x² - 4a²x + (a⁴ - b⁴) = 0
★Provide Correct answer with detailed explanation.
★ No spamming.
★ No irrelevant answers.
★ No offensive meeting links.
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Verified answer
[tex]\large\underline{\sf{Solution-}}[/tex]
Given quadratic equation is
[tex]\sf \: {4x}^{2} - {4a}^{2}x + ( {a}^{4} - {b}^{4}) = 0 \\ \\ [/tex]
[tex]\sf \: {4x}^{2} - {4a}^{2}x + {a}^{4} - {b}^{4} = 0 \\ \\ [/tex]
[tex]\sf \: ({4x}^{2} - {4a}^{2}x + {a}^{4}) - {b}^{4} = 0 \\ \\ [/tex]
[tex]\sf \: [ \: {(2x)}^{2} - {4a}^{2}x + {( {a}^{2} )}^{2} \: ]- {b}^{4} = 0 \\ \\ [/tex]
[tex]\sf \: [ \: {(2x)}^{2} - 2 \times x \times {a}^{2} + {( {a}^{2} )}^{2} \: ]- {b}^{4} = 0 \\ \\ [/tex]
[tex]\sf \: {(2x - {a}^{2} )}^{2} - {b}^{4} = 0 \\ \\ [/tex]
[tex]\sf \: {(2x - {a}^{2} )}^{2} - {( {b}^{2} )}^{2} = 0 \\ \\ [/tex]
[tex]\sf \: (2x - {a}^{2} + {b}^{2})(2x - {a}^{2} - {b}^{2}) = 0 \\ \\ [/tex]
[tex]\sf \: 2x - {a}^{2} + {b}^{2} = 0 \: \: \: or \: \: \: 2x - {a}^{2} - {b}^{2} = 0 \\ \\ [/tex]
[tex]\sf \: 2x = {a}^{2} - {b}^{2} \: \: \: or \: \: \: 2x = {a}^{2} + {b}^{2} \\ \\ [/tex]
[tex]\bf\implies \:x = \dfrac{ {a}^{2} - {b}^{2} }{2} \: \: \: or \: \: \: x = \dfrac{ {a}^{2} + {b}^{2} }{2} \\ \\ [/tex]
[tex]\rule{190pt}{2pt}[/tex]
Identities Used :-
[tex]\boxed{ \sf{ \: {(x - y)}^{2} = {x}^{2} - 2xy + {y}^{2} \: }} \\ \\ [/tex]
[tex]\boxed{ \sf{ \: {x}^{2} - {y}^{2} \: = \: (x + y)(x - y) \: }} \\ \\ \\ [/tex]
[tex]\rule{190pt}{2pt}[/tex]
[tex] {{ \bf{Additional\:Information}}}[/tex]
Nature of roots :-
Let us consider a quadratic equation ax² + bx + c = 0, then nature of roots of quadratic equation depends upon Discriminant (D) of the quadratic equation.
Three cases arise :-
If Discriminant, D > 0, then roots of the equation are real and unequal.
If Discriminant, D = 0, then roots of the equation are real and equal.
If Discriminant, D < 0, then roots of the equation are unreal or complex or imaginary.
Where,
Discriminant, D = b² - 4ac