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solve the equations 2x²-3×+1=0 and x²+2×-3-=0 then make explanation to the solution how did you solve the equations​

Sagot :

AnonymousBetaidn

Answer:

x_1 = 1/2 , x_2 = 1

Step-by-step explanation:

2x^2 - 3x + 1 =0

x^2 - (3/2)x + 1/2 = 0

next is let x_1 and x_2 be the roots of the equation

let x_1 = -b/2 - u and x_2 = -b/2 + u

it is known that

(x_1)(x_2) = c

if a=1

in ax^2 + bx + c = 0

then lets substitite with the values we let in

(-b/2 - u)(-b/2 + u) = c

(3/4 - u)(3/4 +u) = 1/2

9/16 - u^2 = 1/2

u^2 = 9/16 - 8/16

u = +/- 1/4

then find x_1 and x_2 with the values you got

x_1 = -b/2 - u

=(3/2)(1/2) - 1/4

=3/4 - 1/4

x_1=1/2

x_2 = -b/2 + u

=(3/2)(1/2) + 1/4

=3/4 + 1/4

x_2=1

then for x^2 + 2x - 3 = 0

with the form of ax^2 + bx + c = 0

Let x_1 = -b/2 - u , x_2 = -b/2 + u

it is known that

(x_1)(x_2) = c

then sub

(-b/2 - u)(-b/2 + u ) = c

(-2/2 - u)(-2/2 + u) = -3

1 - u^2 = -3

u^2 = 4

u = +/-2

then find the values for x_1 amd x_2

x_1 = - b/2 - u

=-2/2 - 2

= -1 - 2

x_1 = -3

x_2 = -b/2 + u

=-2/2 + 2

=-1 + 2

x_2 = 1