Answer to Question #285509 in Algebra for bookaddict

Question

find the values of k for which the following equations have no real roots.

(a) kx^2-4x+8=0

(b) 3x^2+5x+k+1=0

(c) 2x^2+8x-5=kx^2

(d) 2x^2+k=3(x-2)

(e) kx^2+2kx=4x-6

(f) kx^2+kx=3x-2

Answer 1

(a)

$$kx^2-4x+8=0$$

$k\not=0$

$$D=(-4)^2-4(k)(8)<0$$

$$16-32k<0$$

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

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

(b)

$$3x^2+5x+k+1=0$$

$$D=(5)^2-4(3)(k+1)<0$$

$$13-12k<0$$

$$k>\dfrac{13}{12}$$

$$k\in (\dfrac{13}{12}, \infin)$$

(c)

$$(2-k)x^2+8x-5=0$$

$k\not=2$

$$D=(8)^2-4(2-k)(-5)<0$$

$$16+10-5k<0$$

$$26-5k<0$$

$$k>\dfrac{26}{5}$$

$$k\in (\dfrac{26}{5}, \infin)$$

(d)

$$2x^2-3x+6+k=0$$

$$D=(-3)^2-4(2)(6+k)<0$$

$$9-48-8k<0$$

$$8k>-39$$

$$k>-\dfrac{39}{8}$$

$$k\in (-\dfrac{39}{8}, \infin)$$

(e)

$$kx^2+2kx-4x+6=0$$

$k\not=0$

$$D=(2k-4)^2-4(k)(6)<0$$

$$4k^2-16k+16-24k<0$$

$$k^2-10k+4<0$$

$$k^2-10k+25-21<0$$

$$(k-5)^2<21$$

$$5-\sqrt{21}<k<5+\sqrt{21}$$

$$k\in(5-\sqrt{21},5+\sqrt{21})$$

(f)

$$kx^2+kx=3x-2$$

$k\not=0$

$$kx^2+(k-3)x+2=0$$

$$D=(k-3)^2-4(k)(2)<0$$

$$k^2-6k+9-8k<0$$

$$k^2-14k+49<40$$

$$(k-7)^2<40$$

$$7-2\sqrt{10}<k<7+2\sqrt{10}$$

$$k\in (7-2\sqrt{10},7+2\sqrt{10})$$