Question #285507

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


1
Expert's answer
2022-01-10T16:50:21-0500

(a)


x2+8x+3k=0x^2+8x+3-k=0

D=(8)24(1)(3k)>0D=(8)^2-4(1)(3-k)>0

4(13+k)>04(13+k)>0

k>13k>-13

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

(b)


2x25x+k4=02x^2-5x+k-4 =0

D=(5)24(2)(k4)>0D=(-5)^2-4(2)(k-4)>0

578k>057-8k>0

k<578k<\dfrac{57}{8}

k(,578)k\in(-\infin, \dfrac{57}{8})

(c)


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

k0k\not=0

D=(4)24(k)(2)>0D=(-4)^2-4(k)(2)>0

168k>016-8k>0

k<2k<2

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


(d)


kx2+2(k1)x+k=0kx^2+2(k-1)x+k=0

k0k\not=0

D=(2(k1))24(k)(k)>0D=(2(k-1))^2-4(k)(k)>0

12k>01-2k>0

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

k(,0)(0,12)k\in(-\infin, 0)\cup(0, \dfrac{1}{2})

(e)


2x2=2(x1)+k2x^2=2(x-1)+k

x2x+(1k2)=0x^2-x+(1-\dfrac{k}{2})=0

D=(1)24(1)(1k2)>0D=(-1)^2-4(1)(1-\dfrac{k}{2})>0

3+2k>0-3+2k>0

k>32k>\dfrac{3}{2}

k(32,)k\in (\dfrac{3}{2}, \infin)

(f)


kx2+(2k5)x+k1=0kx^2+(2k-5)x+k-1=0

k0k\not=0


D=(2k5)24(k)(k1)>0D=(2k-5)^2-4(k)(k-1)>0

4k220k+254k2+4k>04k^2-20k+25-4k^2+4k>0

16k+25>0-16k+25>0

k<2516k<\dfrac{25}{16}

k(,0)(0,2516)k\in(-\infin, 0)\cup(0, \dfrac{25}{16})


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