Question

What are the set of values for which this equation has real distinct roots?

`2x^2 + 3kx +k=0`

Answer 1

`k<0" or "k>8/9`

Explanation:

`"to determine the nature of the roots use the "color(blue)"discriminant"`

`•color(white)(x)Delta=b^2-4ac`

`• " If "Delta>0" then real distinct roots"`

`• " If "Delta=0" then real and equal roots"`

`• " If "Delta<0" then complex roots"`

`"here "Delta>0" is required"`

`2x^2+3kx+k=0larrcolor(blue)"is in standard form"`

`"with "a=2,b=3k" and "c=k`

`rarrDelta=(3k)^2-(4xx2xxk)=9k^2-8k`

`"rArr9k^2-8k>0`

`"the left side is a quadratic with positive leading"`
`"coefficient and zeros at "k=0" and "k=8/9`
graph{9x^2-8x [-10, 10, -5, 5]}

`"Thus it is positive when "k<0" or "k>8/9`

`k in(-oo,0)uu(8/9,oo)`

Answer 2

`k<0` `" , "` `k>8/9`

Explanation:

`2x^2+3kx+k=0`
is a quadratic equation

and to find the roots of `x` of quadratic equations

`color(green)"Example :"` `color(blue)(ax^2+bx+c=0)`

we use the following formula

`color(blue)(x=(-b+-sqrt(b^2-4ac))/(2a)`
`color(blue)("where the term "(b^2-4ac) " is the discriminant"`

so in the quadratic equation

If the discriminant
`b^2-4ac>0` `rarr`the equation has two real solutions .
`b^2-4ac<0` `rarr`the equation has no real solutions .
`b^2-4ac=0` `rarr`the equation has one real solution .

`2x^2+3kx+k=0`

`a=2` `" , "` `b=+3k` `" , "` `c=k`

Substitute in the discriminant

`9k^2-(4)(2)(k)`

so in order to get the real distinct roots of the function

`color(blue)("discriminant">0`

`9k^2-8k>0`

`k(9k-8)>0`

`k<0` `" , "` `k>8/9`