IIT JEE , AIEEE Forum - Mathematics , Algebra - please solve it !!! [admin]: quadratic equation numerical

vishnu_srivastava

Find the set of values for which the equation

(x²+x)² +a(x²+x)+4=0 has 4 distinct roots.

hints:

use by taking x²+x=t

also

t²+a+4=0 {use quadratic to find restrictions on t }

krish

For the solution of the above question ,
the discriminant of x^2+x-t is put on trial and must be >or = 0 and therefore putting the value of t interms of a , u will get a>=65/4
if u find any mistake , reply

puneet

hii

U have asked for 4 distinct values and not 4 distinct real values .. rite ??

Now put x² + x = t .. so we get

t²+ at + 4 = 0

solving we get t = (-a +

a²- 16)/2 , (-a -

a²- 16)/2

Now t shud have 2 distinct values so that further we can take 2 quadratic quations ..

So a

4, -4 ...... (1)

Now we put x² + x = (-a +

a²- 16)/2 , (-a -

a²- 16)/2

if (-a -

a²- 16)/2 = -1/4 or (-a +

a²- 16)/2 = -1/4 (B² - 4AC = 0)

then again there will be one solution for that equation as well and hence we will not get 4 distinct solutions ..

So solving we get a

65/4 ... (2)

Combining (1) and (2) we get ..

a

4, -4 , 65/4

cheers

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