5.      Solve the radical equation .

Square both side of the equation

(?2x-5+1)2=(x-1)2

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=2x-5+1=(x-1)2

2x-4=(x-1)(x+1)

2x-4=(x2+x-x-1)

2x-4=(x2-1)

2x-4-x2+1=0

2x-x2-3=0

-x2+2x-3=0

Multiply by (-1)

-1(-x2+2x-3) =0(-1)

X2-2x+3=0

-b±?b2-4ac/2a

-(-2)± ?-22-(4.1.3)/2.1

=2+?4-(12)/2 and 2-?4-(12)/2

=2+?-8/2 and 2-?-8/2

Here the solution is indeterminate.

But if I assume that the square does not cover the +1 then I can redo the equation as this.

5. Solve the radical equation .

Subtract 1 from both equations

(?2x-5)+1-1= x-1-1

Square both side of the equation

(?2x-5)2=(x-1-1)2

= 2x-5=(x-2)2

=2x-5=(x-2) (x+2)

2x-5=(x2-4)

=x2-4-2x+5=0

=x2-2x+1=0

Solving the equation through fraction

(x-1)(x-1)=0

So (x-1) =0

X=1for both equation

Therefore x=1

6. Use the quadratic formula to solve the equation

A=2, b=-5. C=-1

-b±?b2-4ac/2a

-(-5)± ?(-52)-(4.2.-1)/2.2

=5±?25-(-8)/4

=5±?25+8/4

=5±?33/4

=5+?33/4 or 5-?33/4

=+2.686 and -0.186

Therefore x=+2.686 or -0.186.

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