Help me answer this question for 5.b)

The provement is below

Step-by-step explanation:

z^(1/2)=x^(1/2)+y^(1/2)  =>  (z^(1/2))^2=  (x^(1/2)+y^(1/2))^2

=> z=x+y+2*x^(1/2)*y^(1/2) => z-x-y= 2*x^(1/2)*y^(1/2)

=> (z-x-y)^2= (2*x^(1/2)*y^(1/2) )^2 => (z-x-y)^2=4*x*y           (1)

Pls note that (z-x-y)^2= ((-1)*(-1)*(z-x-y))^2= ((-1)*(x+y-z))^2= (-1)^2*(x+y-z)^2=

=(x+y-z)^2

So  (z-x-y)^2= (x+y-z)^2 !!! Substitute in (1) (z-x-y)^2 by (x+y-z)^2   and will get

the required equality  (x+y-z)^2=4*x*y

step-by-step explanation:

x = ±sqrt(10)

step-by-step explanation:

5 - 2x² = -15

subtract 5 from each side

5-5 - 2x² = -15-5

-2x^2   = -20

divide by -2 on each side

-2x^2   /-2= -20/-2

x^2 =10

take the square root on each side

sqrt(x^2 )=±sqrt(10)

x = ±sqrt(10)

check:

5 - 2x² = -15

+sqrt(10)

5 - 2(sqrt(10))² = -15

5 -2(10) = -15

5 -20 = -15

-15 = -15

-sqrt(10)

5 - 2(-sqrt(10))² = -15

5 -2(10) = -15

5 -20 = -15

-15 = -15