Solve for x. sqrt(4x+5)-sqrt(x-1)=3
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Square root of pkchu.
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[quote]Solve for x. sqrt(4x+5)-sqrt(x-1)=3[/quote] Sqrt= pnutbutr pnutbutr(4x+5)-pnutbutr(x-1)=3 4xpnutbutr+xpnutbutr= 5xpnutbutr 5pnutbutr-1pnutbutr=4pnutbutr 5xpnutbutr/4pnutbutr=3 3=3
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Calculus Tutor here. There's a lot of people in the thread doing this wrong; you guys need to remember how to square equations. Begin. sqrt(4x + 5) - sqrt(x - 1) = 3 Add sqrt(x - 1) to both sides. sqrt(4x + 5) = 3 + sqrt(x - 1) Square both sides. Note that this looks like: (sqrt(4x + 5))^2 = (3 + sqrt(x - 1))^2 Not... (sqrt(4x + 5))^2 = (3)^2 + (sqrt(x - 1))^2 This is the mistake that everyone else in the thread has made at the time of writing this post. Continuing... (sqrt(4x + 5))^2 = (3 + sqrt(x - 1))^2 Simplifying on the left... 4x + 5 = (3 + sqrt(x - 1))^2 Foiling out the right... 4x + 5 = 9 + 6*sqrt(x - 1) + (x - 1) Simplify on the right... 4x + 5 = 8 + 6*sqrt(x - 1) + x Subtract x and 8 from both sides... 3x - 3 = 6*sqrt(x - 1) Square both sides... (3x - 3)^2 = (6*sqrt(x - 1))^2 Simplify the right... (3x - 3)^2 = 36*(x - 1) Foil out the left... 9x^2 - 18x + 9 = 36*(x - 1) Factor out a 9 on the left... 9(x^2 - 2x + 1) = 36*(x - 1) Divide both sides by 9. x^2 - 2x + 1 = 4(x - 1) Simplify on the right... x^2 - 2x + 1 = 4x - 4 Subtract 4x from both sides and add 4 to both sides... x^2 - 6x + 5 = 0 Now we need to factor the quadratic. I need two numbers that multiply together to 5, and add together to -6. -5 and -1 seem like perfect candidates. Completing the factoring process we get: (x - 5)(x - 1) = 0 Ba-BOMB! x = 5. x = 1. Any questions?
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use wolfram alpha.
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google is your lover
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Damnit tyrone keep it together
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potato
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Square it all to get: 4x-1-x-1=9 Then: 3x-2=9 3x=11 x=11/3
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Square both sides so you get (4x-1)-(x-1)=9 Open parenthesis 4x-1-x+1=9 As the ones cancel each other out 4x-x=9 Subtraction 3x=9 Divide both sides by 3 (3x)/3=9/3 You have solved for x x=3
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sqrt(4x-1) - sqrt(x-1) = 3 Square everything. (4x-1) - (x-1) = 9 4x-1 - x+1 = 9 3x = 9 x = 3
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I should know this.
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sqrt is sqaure root right?