Question

if x is real and 4y^+4xy+x+6=0 then the values of x for which y is real​

Answer 1

hope this answer helps you and plz mark it as the brainliest

Answer 2

Given:

$4y^2+4xy+x+6=0$

$x$ is real.

To Find:

Value of $x$ for which $y$ is real.

Solution:

We are given a quadratic equation $4y^2+4xy+x+6=0$.

A quadratic equation $ax^2+bx+c=0$ can have the following pair of roots

• Real and distinct.
• Real and equal.
• Complex conjugates.

And the type of root is decided by the discriminant, $D=b^2-4ac$.

If $D > 0$, roots are real and distinct,

If $D=0$, roots are real and equal,

And if $D < 0$, roots are complex conjugates.

So for $y$ to be real, the discriminant of the given equation will have to be greater than or equal to $0$.

    $D\geq 0$

⇒ $(4x)^2-4(4)(x+6)\geq 0$

⇒ $16x^2\geq 16x^2+96$

But this condition is not possible.

Hence, for no value of $x$, $y$ can have real value.