Divide 20 into two parts such that three times the square of one part exceeds the other part by 10
Answers
Answered by
6
Let two parts be x and y
So , x + y =20 -(1)
Now,
3x^2 = y + 10
y = 3x^2 - 10
Put in (1)
x + 3x^2 -10 = 20
x + 3x^2 = 30
3x^2 + x - 30 = 0
Solve quadratic equation
x = (-b +- √b^2 - 4ac) / 2a
x = -1 +- √1 - 4(3)(-30) / 2(3)
= -1 +- √1 + 360 / 6
= -1 +- √361 / 6
= -1 + 19 /6 , -1 -19/6
= 3 , -20/6=-10/3
Let x = 3
Put in equation (1)
3 + y = 20
y = 17
If x = -10/3
Put in equation (1)
-10/3 + y = 20
y = 20 + 10/3 = 23.3
So , x + y =20 -(1)
Now,
3x^2 = y + 10
y = 3x^2 - 10
Put in (1)
x + 3x^2 -10 = 20
x + 3x^2 = 30
3x^2 + x - 30 = 0
Solve quadratic equation
x = (-b +- √b^2 - 4ac) / 2a
x = -1 +- √1 - 4(3)(-30) / 2(3)
= -1 +- √1 + 360 / 6
= -1 +- √361 / 6
= -1 + 19 /6 , -1 -19/6
= 3 , -20/6=-10/3
Let x = 3
Put in equation (1)
3 + y = 20
y = 17
If x = -10/3
Put in equation (1)
-10/3 + y = 20
y = 20 + 10/3 = 23.3
Similar questions