Question

2y2-2==0 (5-2t)*(7+5t)=0

Answer 1

2y^2-2=0
2y^2=2
y^2=1
y1=1
y2=-1


(5-2t)*(7+5t)=0
5-2t=0
2t=5
t=5/2= 2 1/2

или

7+5t=0
5t=-7
t=-7/5= -1 2/5

Answer 2

To solve the equations, set each expression equal to zero and then find the values of 't' that satisfy each equation.

1. 2y^2 - 2 = 0:

To solve this quadratic equation for 'y,' we can factor it: 2y^2 - 2 = 0 2(y^2 - 1) = 0

Now, we have a difference of squares: 2(y - 1)(y + 1) = 0

Now, set each factor equal to zero and solve for 'y': y - 1 = 0 --> y = 1 y + 1 = 0 --> y = -1

So, the solutions for this equation are y = 1 and y = -1.

2. (5 - 2t)(7 + 5t) = 0:

To solve this equation, we can use the zero product property, which states that if the product of two factors is zero, then at least one of the factors must be zero.

So, set each factor equal to zero and solve for 't': 5 - 2t = 0 --> -2t = -5 --> t = 5/2

7 + 5t = 0 --> 5t = -7 --> t = -7/5

The solutions for this equation are t = 5/2 and t = -7/5.

So, the solutions for the two equations are as follows:

For 2y^2 - 2 = 0: y = 1 and y = -1 For (5 - 2t)(7 + 5t) = 0: t = 5/2 and t = -7/5