The answer is: " x = 5, and -1 " .
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Step-by-step explanation:
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Given:
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(x − 2) = √(2x + 1) ;
→ "Square" each side of the equation; that is, raised EACH SIDE of the equation to the power of "2" ; to get rid of the "radical" ; as follows:
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[(x - 2)]² = [ √(2x + 1) ] ² ;
to get:
→ (x - 2)² = (2x - 1) ;
Expand the " (x - 2)² " ;
→ (x - 2)(x -2) = ?
NOTE: (a + b)(c + d) = ac + ad + bd .
→ = (x * x) + (x * -2) + (-2 * x) + (-2 * -2) ;
= (x²) + (-2x) + (-2x) + (4) ;
= x² − 2x − 2x + 4 ;
= x² − 4x + 4 ;
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Now, rewrite the entire question:
→ (x − 2)² = (2x − 1) ;
→ as follows:
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→ " x² − 4x + 4 = 2x − 1 " ;
Now, subtract "2x" from each side of the equation;
& add "1" to each side of the equation:
→ " x² − 4x + 4 − 2x + 1 = 2x − 1 − 2x + 1 " ;
to get:
→ " x² − 6x + 5 = 0 " ; Solve for "x" ;
Let us if : " x² − 6x + 5 " can be factors:
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What factors of "positive 5" ; add up to "-6" ;
-5 ,-1 ; → -5 + (-1) = - 6; YES.
5, 1 ; → 5 + 1 = 6 ; NO.
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So, we have: " (x − 5) (x − 1) = 0 " .
x = 5, and -1 .
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