Answer to Question #285507 in Algebra for bookaddict

Question

find the values of k for which the following equations have two distinct roots.

(a) x^2+8x+3=k

(b) 2x^2-5x =4-k

(c) kx^2-4x+2=0

(d) kx^2+2(k-1)x+k=0

(e) 2x^2=2(x-1)+k

(f) kx^2+(2k-5)x=1-k

Answer 1

(a)

$$x^2+8x+3-k=0$$

$$D=(8)^2-4(1)(3-k)>0$$

$$4(13+k)>0$$

$$k>-13$$

$$k\in( -13, \infin)$$

(b)

$$2x^2-5x+k-4 =0$$

$$D=(-5)^2-4(2)(k-4)>0$$

$$57-8k>0$$

$$k<\dfrac{57}{8}$$

$$k\in(-\infin, \dfrac{57}{8})$$

(c)

$$kx^2-4x+2=0$$

$k\not=0$

$$D=(-4)^2-4(k)(2)>0$$

$$16-8k>0$$

$$k<2$$

$$k\in(-\infin, 0)\cup(0, 2)$$

(d)

$$kx^2+2(k-1)x+k=0$$

$k\not=0$

$$D=(2(k-1))^2-4(k)(k)>0$$

$$1-2k>0$$

$$k<\dfrac{1}{2}$$

$$k\in(-\infin, 0)\cup(0, \dfrac{1}{2})$$

(e)

$$2x^2=2(x-1)+k$$

$$x^2-x+(1-\dfrac{k}{2})=0$$

$$D=(-1)^2-4(1)(1-\dfrac{k}{2})>0$$

$$-3+2k>0$$

$$k>\dfrac{3}{2}$$

$$k\in (\dfrac{3}{2}, \infin)$$

(f)

$$kx^2+(2k-5)x+k-1=0$$

$k\not=0$

$$D=(2k-5)^2-4(k)(k-1)>0$$

$$4k^2-20k+25-4k^2+4k>0$$

$$-16k+25>0$$

$$k<\dfrac{25}{16}$$

$$k\in(-\infin, 0)\cup(0, \dfrac{25}{16})$$