# Math Help - Simple inequality

1. ## Simple inequality

The question is:

Given that the equation kx^2 - 4x + (k-3) = 0 has real roots, show that k^2 - 3k - 4 < 0.
Find the range of values of k satisfying this inequality.

b^2 - 4ac = 0

(-4)^2 - 4(k)(k-3) = 0
16 - 4k^2 + 12k = 0
(/4 4 - k^2 + 3k = 0
(Rearranged) k^2 - 3k - 4 = 0
(Factorized) (k-4)(k+1) = 0
k = 4, -1

I'm guessing that the rest involves the solution of an inequality but what would I do here?

2. Hello, db5vry!

You had a right approach, but we want an inequality . . .

Given that the equation $kx^2 - 4x + (k-3) \:= \:0$ has real roots,

(a) show that: $k^2 - 3k - 4 \:< \:0$

(b) Find the range of values of $k$ satisfying this inequality.
(a) A quadratic has two real roots if its discriminant is positive: . $b^2 - 4ac \:> \:0$

. . So we have: . $(\text{-}4)^2 - 4(k)(k-3) \:>\:0 \quad\Rightarrow\quad -4k^2 + 12k + 16 \:>\:0$

. . Divide by -4: . $\boxed{k^2 - 3k - 4 \:<\:0}$

(b) We have: . $k^2 - 3k - 4 \:<\:0$

. . Factor: . $(k+1)(k-4) \:<\:0$

The product of two factors is negative if they have opposite signs.
So there are two cases:

$[1]\;\;\begin{array}{ccc}k+1 \:< 0 & \Rightarrow & k \:<\:\text{-}1 \\ k-4\:>\:0 & \Rightarrow & k \:>\:4 \end{array} \quad\Rightarrow\quad k < \text{-1}\,\text{ and }\,k > 4\quad\hdots\;\;\text{Impossible}$

$[2]\;\;\begin{array}{ccc}k+1 \:>\:0 & \Rightarrow & k \:>\:\text{-}1 \\ k-4 \:<\:0 & \Rightarrow & k \:<\:4 \end{array}\quad\Rightarrow\quad k > \text{-}1\,\text{ and }\,k < 4\quad\hdots\;\;\text{Yes!}$

Answer: . $\boxed{\text{-}1 \:<\:k\:<\:4}$