SOLUTION: The directions are: Solve over the complex numbers
The problems are b-3 sqrt of b -10=0 (b minus 3 times the square root of b minus 10 equal 0.
and 2x-9 sqrt x + 4=0 (2x minus
Algebra.Com
Question 566000: The directions are: Solve over the complex numbers
The problems are b-3 sqrt of b -10=0 (b minus 3 times the square root of b minus 10 equal 0.
and 2x-9 sqrt x + 4=0 (2x minus 9 times the square root of x plus 4)
The title of the paper is the Discriminant: equations in quadratic form. Since there is no squared number in these I don't know how to begin doing them.
Thank you!
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
The problems are b-3 sqrt of b -10=0 (b minus 3 times the square root of b minus 10 equal 0.
---------------------------
b - 3sqrt(b) - 10 = 0
Note: This is a quadratic with variable sqrt(b)
Factor;
(sqrt(b)-5)(sqrt(b)+2) = 0
sqrt(b) = 5 or sqrt(b) = -2
b = 25 or b = 4
Note: The b = 4 solution may be extraneous
as sqrt(b) cannot be negative.
====================================
and 2x-9 sqrt x + 4=0 (2x minus 9 times the square root of x plus 4)
2x - 9sqrt(x) + 4 = 0
---
2x-8sqrt(x)-sqrt(x) + 4 = 0
----
2sqrt(x)(sqrt(x)-4)-(sqrt(x)-4) = 0
-----
(sqrt(x)-4)(2sqrt(x)-1) = 0
sqrt(x) = 4 or sqrt(x) = 1/2
x = 16 or x = 1/4
========================
Cheers,
Stan H.
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