Answer to Question #163840 in Algebra for Xavier

Question

15. If x is real and: px² + 2px - 3 < 0 then the possible values of p are ?

 16.if x is real the set of values of x² + 4x - 2 = 0 is ?

Answer 1

1).

$px^2 +2px-3\lt0$

if x is real then, solution of x is:

$x=\frac{-2p\pm\sqrt{(2p)^2-4*p*(-3)}}{2p}=\frac{-2p\pm\sqrt{4p^2+12p}}{2p}$

from the above equation, for the real solution of x, there are some restriction to be follow as:

$p\ne0$ and

value of p lies in set as: $4p^2+12p\ge0$

$\implies4p^2\ge-12p$

$\implies p\ge-3$

thus, the possible values of p are$[-3, ∞)-(0)$.

2).

$x^2+4x-2=0$

if x is real then the set of values of equation $x^2+4x-2$ is:

$x=\frac{-4\pm\sqrt{4^2-4*1*(-2)}}{2*1}=\frac{-4\pm\sqrt{16+8}}{2}=-2\pm\sqrt6$

thus, the set of values of x²Â + 4x - 2 = 0 are $[( -2-\sqrt6),(-2+\sqrt6)]$