Mister Exam

Other calculators

x^2-y^2=0 equation

The teacher will be very surprised to see your correct solution 😉

v

Numerical solution:

Do search numerical solution at [, ]

The solution

You have entered [src]
 2    2    
x  - y  = 0
x2y2=0x^{2} - y^{2} = 0
Detail solution
This equation is of the form
a*x^2 + b*x + c = 0

A quadratic equation can be solved
using the discriminant.
The roots of the quadratic equation:
x1=Db2ax_{1} = \frac{\sqrt{D} - b}{2 a}
x2=Db2ax_{2} = \frac{- \sqrt{D} - b}{2 a}
where D = b^2 - 4*a*c - it is the discriminant.
Because
a=1a = 1
b=0b = 0
c=y2c = - y^{2}
, then
D = b^2 - 4 * a * c = 

(0)^2 - 4 * (1) * (-y^2) = 4*y^2

The equation has two roots.
x1 = (-b + sqrt(D)) / (2*a)

x2 = (-b - sqrt(D)) / (2*a)

or
x1=y2x_{1} = \sqrt{y^{2}}
x2=y2x_{2} = - \sqrt{y^{2}}
Vieta's Theorem
it is reduced quadratic equation
px+q+x2=0p x + q + x^{2} = 0
where
p=bap = \frac{b}{a}
p=0p = 0
q=caq = \frac{c}{a}
q=y2q = - y^{2}
Vieta Formulas
x1+x2=px_{1} + x_{2} = - p
x1x2=qx_{1} x_{2} = q
x1+x2=0x_{1} + x_{2} = 0
x1x2=y2x_{1} x_{2} = - y^{2}
The graph
Sum and product of roots [src]
sum
-re(y) - I*im(y) + I*im(y) + re(y)
(re(y)iim(y))+(re(y)+iim(y))\left(- \operatorname{re}{\left(y\right)} - i \operatorname{im}{\left(y\right)}\right) + \left(\operatorname{re}{\left(y\right)} + i \operatorname{im}{\left(y\right)}\right)
=
0
00
product
(-re(y) - I*im(y))*(I*im(y) + re(y))
(re(y)iim(y))(re(y)+iim(y))\left(- \operatorname{re}{\left(y\right)} - i \operatorname{im}{\left(y\right)}\right) \left(\operatorname{re}{\left(y\right)} + i \operatorname{im}{\left(y\right)}\right)
=
                  2
-(I*im(y) + re(y)) 
(re(y)+iim(y))2- \left(\operatorname{re}{\left(y\right)} + i \operatorname{im}{\left(y\right)}\right)^{2}
-(i*im(y) + re(y))^2
Rapid solution [src]
x1 = -re(y) - I*im(y)
x1=re(y)iim(y)x_{1} = - \operatorname{re}{\left(y\right)} - i \operatorname{im}{\left(y\right)}
x2 = I*im(y) + re(y)
x2=re(y)+iim(y)x_{2} = \operatorname{re}{\left(y\right)} + i \operatorname{im}{\left(y\right)}
x2 = re(y) + i*im(y)