#### The equation x^2 + kx + (k + 3) = 0, where k is a constant, has different real roots.

(a) Show that k^2\ -\ 4k\ -\ 12\ >0

(b) Find the set of possible values of k.

Anonymous

0

(a) Show that k^2\ -\ 4k\ -\ 12\ >0

(b) Find the set of possible values of k.

Krishna

0

Step 1: Note-down the given quadratic equation and compare it with the standard form ax^2 + bx + c

EXAMPLE: 2x^2 - 3x - (k+1)

ax^2 + bx + c

Where a = 2, b = -3 and c = -(k+1)

Step 2: Look the hints given in the question according to that hints choose the discrimination( b^2-4ac ) of the equation.

HINTS:1) b^2 - 4ac > 0 (Positive), there are 2 real solutions

2) b^2-4ac = 0 (Zero), there is one real solution

3) b^2-4ac < 0 (Negative), there are 2 complex solutions or no real roots.

EXAMPLE: HINT: Has no real roots

So we have to take the discrimination of b^2-4ac < 0

Step 3: Substitute all the values of a, b and c in the suitable discrimination.

EXAMPLE: [math] (-3)^2 - 4 [2][-(k+1)] < 0[/math]

Step 4: Simplify the inequality to find the unknown value.

NOTE: Apply inverse operations on both sides, for linear equations.

Apply the factorization method for the quadratic equation.

EXAMPLE: 9 + 8k + 8 <0

8k+17<0

8k < -17

k\ <\ \frac{-17}{8}