Solution set for quadratic equation: (a^2 +a +1)x^2 + (2a - 3)x + a - 5 =0

[GuidingLight]

find all values of a for which one root of the equation is more than 1, and another is less than 1.

(a^2 +a +1)x^2 + (2a - 3)x + a - 5 =0


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Equation has two roots so D>0

D = B^2 - 4AC

B = 2a - 3
A = a^2 + a + 1
C = a - 5

D = (2a - 3)^2 - 4(a^2 + a + 1)(a -5) = -4a^3 + 20a^2 + +4a +29 > 0

That's where i got stuck, can this be simplified? Or am I doing something wrong?

 

[Deleted member 4993]

find all values of a for which one root of the equation is more than 1, and another is less than 1.

(a^2 +a +1)x^2 + (2a - 3)x + a - 5 =0


__________________________________

Equation has two roots so D>0

D = B^2 - 4AC

B = 2a - 3
A = a^2 + a + 1
C = a - 5

D = (2a - 3)^2 - 4(a^2 + a + 1)(a -5) = -4a^3 + 20a^2 + +4a +29 > 0

That's where i got stuck, can this be simplified? Or am I doing something wrong?

let

P = a^2 + a + 1 → >0 for all values of 'a' (Why?)

Q = 2a - 3

R = a - 5

Since, one root of the equation is more than 1, let

[Q + √(Q^2 - 4PR)]/(2P) > 1

[Q + √(Q^2 - 4PR)] > 2P

√(Q^2 - 4PR) > 2P - Q

Q^2 - 4PR > 4P^2 + Q^2 - 4PQ

4P(Q-R) > 4P^2

Q - R > P

a + 2 > a^2 + a + 1

a^2 < 1

Continue.... also check the other condition ([Q - √(Q^2 - 4PR)]/(2P) <1)

 

[GuidingLight]

[Q + √(Q^2 - 4PR)]/(2P) > 1

Q should be with minus

[-Q + √(Q^2 - 4PR)]/(2P) > 1

-Q + √(Q^2 - 4PR) > 2P

√(Q^2 - 4PR) > 2P + Q

Q^2 - 4PR > 4P^2 + 4PQ + Q^2

P + R + Q < 0

a^2 + a + 1 + a - 5 + 2a - 3 < 0

a^2 + 4a - 7 < 0

D= 16 - 28 <0 so that has no solution

let's try another condition where [-Q + √(Q^2 - 4PR)]/(2P) < 1

-Q + √(Q^2 - 4PR) < 2P

√(Q^2 - 4PR) < 2P + Q

Q^2 - 4PR < 4p^2 + 4PQ + Q^2

P = a^2 + a + 1 > 0

so:

R + P + Q < 0

a - 5 + a^2 + a + 1 + 2a - 3 < 0

a^2 + 4a - 7 > 0

D = 16+28=44

a1 = (-4 + √44)/2 = -2 + √11

a2 = (-4 - √44)/2 = -2 - √11

 

[Deleted member 4993]

Q should be with minus

[-Q + √(Q^2 - 4PR)]/(2P) > 1

-Q + √(Q^2 - 4PR) > 2P

√(Q^2 - 4PR) > 2P + Q

Q^2 - 4PR > 4P^2 + 4PQ + Q^2

P + R + Q < 0

a^2 + a + 1 + a - 5 + 2a - 3 < 0

a^2 + 4a - 7 < 0

(a+2)^2 < 11 .......................... (1)

D= 16 - 28 <0 so that has no solution

let's try another condition where [-Q + √(Q^2 - 4PR)]/(2P) < 1

-Q + √(Q^2 - 4PR) < 2P

√(Q^2 - 4PR) < 2P + Q

Q^2 - 4PR < 4p^2 + 4PQ + Q^2 → 0 < R + P + Q

P = a^2 + a + 1 > 0

so:

R + P + Q > 0

a - 5 + a^2 + a + 1 + 2a - 3 > 0

a^2 + 4a - 7 > 0

(a + 2)^2 > 11 ......................................(2)

Conditions (1) and (2) are mutually exclusive - thus no solution.

D = 16+28=44

a1 = (-4 + √44)/2 = -2 + √11

a2 = (-4 - √44)/2 = -2 - √11

Good catch

Please observe the corrections in red above.