Radical Equations: How do I solve this?
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Answer:
(B). roots
Step-by-step explanation:
(a ± b)² = a² ± 2ab + b²
~~~~~~~~
( [tex]\sqrt{4x+5}[/tex] - [tex]\sqrt{x-1}[/tex] )² = ( [tex]\sqrt{x+4}[/tex] )²
( [tex]\sqrt{4x+5}[/tex] )² - 2[tex]\sqrt{(4x+5)(x-1)}[/tex] + ( [tex]\sqrt{x-1}[/tex] )² = ( [tex]\sqrt{x+4}[/tex] )²
4x + 5 - 2[tex]\sqrt{(4x+5)(x-1)}[/tex] + x - 1 = x + 4
2[tex]\sqrt{(4x+5)(x-1)}[/tex] = 4x + 5 + x - 1 - x - 4
[tex]\sqrt{(4x+5)(x-1)}[/tex] = 2x
( [tex]\sqrt{(4x+5)(x-1)}[/tex] )² = ( 2x )²
( 4x + 5 )( x - 1 ) = 4x²
4x² + x - 5 = 4x² ⇒ x = 5