MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Quadratic Formula

For all real ...

Question

For all real value of x and y , $x^{2} + 2 x y + 3 y^{2} - 6 y \geq \frac{- 9}{2}$

Open in App

Solution

Given, $x^{2} + 2 x y + 3 y^{2} - 6 y$

$= (x + y)^{2} + 2 y^{2} - 6 y$

$= (x + y)^{2} + 2 y (y - 3)$

thus it attains its min value when 2y(y-3) attains its min value
at y=3/2 it attains min value with $\frac{- 9}{2}$

at $x = - y (x + y)^{2} -$ is minimum
thus $x^{2} + 2 x y + 3 y^{2} - 6 y \geq - \frac{9}{2}$

Suggest Corrections

thumbs-up

0

Similar questions

Q. The equation to the pair of tangents drawn from (–1, –2) to parabola

x^{2} = 2 y

Q. For real values of x and y, the minimum value of

f (x, y) = x^{2} + 2 x y + 2 y^{2} + 6 y + 10

Minimum value of $x^{2} + 2 x y + 3 y^{2} - 6 x - 2 y$ $x, y, ϵ R$ is

x

y

x^{2} + y^{2} - 2 x y + 4 x + 4 y + 12 = 0

y

can take the values :

Q. The equation of the parabola whose focus lies at the intersection point of the lines

x + y = 3

x - y = 1

and directrix is

x - y + 5 = 0

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Solving QE using Quadratic Formula

MATHEMATICS

Watch in App

explore_more_icon

Explore more

Quadratic Formula

Standard X Mathematics

Join BYJU'S Learning Program

CrossIcon