Question

solve quadratic equation​

Answer 1

Answer :

(0 , 1) or (2/5 , 4/5)

Solution :

• Given : x² + y² = y , x/2 + y = 1

• To find : x , y = ?

We have ,

x² + y² = y ----(1)

Also ,

=> x/2 + y = 1

=> x/2 = 1 - y

=> x = 2(1 - y) -------(2)

Now ,

Putting x = 2(1 - y) in eq-(1) , we get ;

=> [2(1 - y)]² + y² = y

=> 4(1 - 2y + y²) + y² = y

=> 4 - 8y + 4y² + y² = y

=> 4y² + y² - 8y - y + 4 = 0

=> 5y² - 9y + 4 = 0

=> 5y² - 5y - 4y + 4 = 0

=> 5y(y - 1) - 4(y - 1) = 0

=> (y - 1)(5y - 4) = 0

=> y = 1 , 4/5

• If y = 1 , then

=> x = 2(1 - y)

=> x = 2(1 - 1)

=> x = 2•0

=> x = 0

• If y = 4/5 , then

=> x = 2(1 - y)

=> x = 2(1 - 4/5)

=> x = 2•(1/5)

=> x = 2/5

Hence ,

(0 , 1) or (2/5 , 4/5)

Answer 2

Given :

• x² + y² = y
• x / 2 + y = 1

Solution :

= x² + y² = y....... {1 st Equation}

= x / 2 + y = 1 ........ {2 nd Equation}

= x / 2 = 1 - y

= x = 2 ( 1 - y ) ......... {2 nd Equation}

Then, Substituting value of ' x ' in {1 st Equation} :

= x² + y² = y

= [ 2 ( 1 - y ) ]² + y² = y

= [ 4 ( 1 - 2y + y² ) ] + y² = y

= 4 - 8y + 4y² + y² = y

= 5y² - 9y + 4 = 0

= 5y² - 5y - 4y + 4 = 0

= 5y ( y - 1 ) - 4 ( y - 1 ) = 0

= ( y - 1 ) ( 5y - 4 ) = 0

= y = 1 , 4 / 5

Verifying :

When y = 1

= x = 2 ( 1 - y )

= x = 2 ( 1 - 1 )

= x = 2 ( 0 )

= x = 2

When y = 4 / 5

= x = 2 ( 1 - y )

= x = 2 ( 1 - 4 / 5 )

= x = 2 ( 1 / 5 )

= x = 2 / 5

So, ( 2 , 1 ) or ( 2 / 5 , 4 / 5 )