# What is the value of k so that the equation (k^2 + 2)x^2 - 3x^2 + 3x + k + 2 = 0 will have two distinct real roots?

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### 2 Answers

If a quadratic equation ax^2 + bx + c = 0 has two distinct real roots b^2 > 4ac

Here, the equation is (k^2 + 2)x^2 - 3x^2 + 3x + k + 2 = 0

=> x^2(k^2 + 2 - 3) + 3x + (k + 2) = 0

b^2 > 4ac

=> 9 > 4*(k^2 - 1)(k + 2)

=> 9 - 4*(k^2 - 1)(k + 2) > 0

=> 9 - 4*(k^3 - k + 2k^2 - 2) > 0

=> 9 - 4k^3 + 4k - 8k^2 + 8 > 0

=> -4k^3 - 8k^2 + 4k + 17 > 0

=> 4k^3 + 8k^2 - 4k - 17 < 0

I don't see how this can be solved further to yield actual values of k, especially as the expression has complex roots.

**For the equation to have 2 distinct real roots, t****he value of k should satisfy : 4k^3 + 8k^2 - 4k - 16 < 0.**

The equation has 2 distinct roots if and only if it's discriminant is strictly positive:

delta = b^2 - 4ac

a,b,c are the coefficients of the quadratic

delta = 9 - 4*(k^2 + 2 - 3)*(k+2)

delta = 9 - 4(k^2 - 1)(k+2)

delta = 9 - 4k^3 - 8k^2 + 4k + 8

delta = 17 - 4k^3 - 8k^2 + 4k

**The values of k that makes delta to be positive are verifying the inequality 17 - 4k^3 - 8k^2 + 4k>0.**