Question #285509

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


1
Expert's answer
2022-01-10T17:00:43-0500

(a)


kx24x+8=0kx^2-4x+8=0

k0k\not=0

D=(4)24(k)(8)<0D=(-4)^2-4(k)(8)<0

1632k<016-32k<0

k>12k>\dfrac{1}{2}

k(12,)k\in(\dfrac{1}{2},\infin)

(b)


3x2+5x+k+1=03x^2+5x+k+1=0

D=(5)24(3)(k+1)<0D=(5)^2-4(3)(k+1)<0

1312k<013-12k<0

k>1312k>\dfrac{13}{12}

k(1312,)k\in (\dfrac{13}{12}, \infin)

(c)


(2k)x2+8x5=0(2-k)x^2+8x-5=0

k2k\not=2


D=(8)24(2k)(5)<0D=(8)^2-4(2-k)(-5)<0

16+105k<016+10-5k<0

265k<026-5k<0

k>265k>\dfrac{26}{5}

k(265,)k\in (\dfrac{26}{5}, \infin)

(d)


2x23x+6+k=02x^2-3x+6+k=0


D=(3)24(2)(6+k)<0D=(-3)^2-4(2)(6+k)<0

9488k<09-48-8k<0

8k>398k>-39

k>398k>-\dfrac{39}{8}

k(398,)k\in (-\dfrac{39}{8}, \infin)



(e)


kx2+2kx4x+6=0kx^2+2kx-4x+6=0

k0k\not=0


D=(2k4)24(k)(6)<0D=(2k-4)^2-4(k)(6)<0

4k216k+1624k<04k^2-16k+16-24k<0

k210k+4<0k^2-10k+4<0

k210k+2521<0k^2-10k+25-21<0

(k5)2<21(k-5)^2<21

521<k<5+215-\sqrt{21}<k<5+\sqrt{21}

k(521,5+21)k\in(5-\sqrt{21},5+\sqrt{21})

(f)


kx2+kx=3x2kx^2+kx=3x-2

k0k\not=0


kx2+(k3)x+2=0kx^2+(k-3)x+2=0

D=(k3)24(k)(2)<0D=(k-3)^2-4(k)(2)<0

k26k+98k<0k^2-6k+9-8k<0

k214k+49<40k^2-14k+49<40

(k7)2<40(k-7)^2<40

7210<k<7+2107-2\sqrt{10}<k<7+2\sqrt{10}

k(7210,7+210)k\in (7-2\sqrt{10},7+2\sqrt{10})


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