For what values of k will the equation x(14^1/2) + 7 = kx^2 have exactly 2 real solutions? I am stuck so any help would be greatly appreciated!! Also, the 14^1/2 is just supposed to be the sqaure root of fourteen.

1 Answer
Jan 18, 2018

#k > -1/2#

Explanation:

#xsqrt(14)+7=kx^2#

Arrange to get:

#kx^2-xsqrt(14)-7=0#

This is just a quadratic equation of the form #color(red)(ax^2+bx+c)#

With #color(white)(88)color(white)(88)a=k# , #color(white)(88)b=-sqrt(14)color(white)(88)# and #color(white)(88)c =-7#

The discriminant of a quadratic is;

#sqrt(b^2-4ac)#

We know that for:

#sqrt(b^2-4ac)>0# ( The equation has two real and different roots )

#sqrt(b^2-4ac)=0# ( The equation has real and repeated roots )

#sqrt(b^2-4ac)<0# ( The equation has non real or imaginary roots )

We need two real solutions, so we use:

#sqrt(b^2-4ac)>0#

Plugging in our known values:

#sqrt((-sqrt(14))^2-(4(k)(-7)))>0#

#sqrt((14+28k))>0#

Squaring:

#14+28k>0#

#k > -14/28#

#k > -1/2#

We can check this. To make our life easier lets say #k=1#

#x^2-xsqrt(14)-7=0#

By quadratic formula:

#x=(sqrt(14)+sqrt(42))/2color(white)(88)# and #color(white)(88)x=(sqrt(14)-sqrt(42))/2#

These are both real and different, as expected.

Hope it helps.