if the equation x^2+2(a+1)x+9a-5=0 has only negative roots then a satisfies the relation a=6 ..
how?

Question

if the equation x^2+2(a+1)x+9a-5=0 has only negative roots then a satisfies the relation a>=6 ..
how?

hartnn · Answer

find 2 roots using quadratic formula, one is surely negative, 
which is other one ?

hartnn · Answer

-2a -2 -sqrt {...} is negative even if sqrt term becomes greater than -2a-2

samigupta8 · Answer

$$x=\frac{ -2a-2\pm \sqrt{4a^2-24a+24}}{ ? 2}$$

hartnn · Answer

for 
-2a-2 + sqrt {...} term
equate it to 
-2a-2 + sqrt {...} < 0

samigupta8 · Answer

do u want me to aply mid term splitting in that square root term and get the value of a

samigupta8 · Answer

i m getting 20<34 a

raden · Answer

the 3 rules  bellow must be satisfied :
(1) x1 + x2 < 0
(2) x1 * x2 > 0
(3) D >= 0
extra information (known) a >=0

with x1, x2 = roots
D = discriminant

raden · Answer

oppss.. i mean a >=6

samigupta8 · Answer

yaa bt how that's wht we have to prove

raden · Answer

just find solution for a which satisfies that equation. x^2+2(a+1)x+9a-5=0 (1) x1 + x2 < 0 -b/a < 0 -2(a+1) < 0 a + 1 > 0 a > -1

raden · Answer

can you solve for the 2nd case ?
(2) x1 * x2 > 0

samigupta8 · Answer

yaaa sure

samigupta8 · Answer

a>5/9 and a>-1

raden · Answer

a > -1 in the 1st case
yeah, the 2nd is right
how the 3rd case ?

samigupta8 · Answer

4a^2-24a+24>=0

raden · Answer

hmmm, i got different like yours.
4a^2 - 28a + 24>=0, it can simplied be
a^2 - 7a + 6 >=0

samigupta8 · Answer

yaa sorry urs is correct

samigupta8 · Answer

now do we need to take intersection of al of them

raden · Answer

yes, exactly find the intersection of them

samigupta8 · Answer

i got the answer thanks...

raden · Answer

very welcome, dear :)

raden · Answer

intersection :
a > = 6, right ?