#### The equation kx^2 + 4x + (5 - k) = 0, where k is a constant, has 2 different real solutions for x.

(a) Show that k satisfies. k^2 - 5k + 4 = 0

Anonymous

0

(a) Show that k satisfies. k^2 - 5k + 4 = 0

Krishna

0

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

EXAMPLE: kx^2+4x+(5-k)

ax^2 + bx + c

Where a = k, b = 4 and c = (5 - k)

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 2 different real solutions

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](4)^2-4[k][(5-k)]>0[/math]

Step 4: Simplify the inequality to find the quadratic equation

EXAMPLE: 16\ -\ 20k\ +4k^2\ >\ 0

take 4 as common

k^2\ -5k\ +4\ >\ 0

Step 5: Verify that the result(in step 4) is satisfying the given answer or not.