Question

Find the value of k for which the roots are real and equal x^2-2(5+2k)x+3(7+10k)=0​

Answer 1

Answer:

Step-by-step explanation:

x^2(5+2k)+3(7+10k)=0

a=1

b=-2(5+2k)

c=3(7+10k)

D=0. Or b^2-4ac=0

[-2(5+2k)]-4(1)[3(7+10k)]=0

4(5+2k)^2-12(7+10k)=0

4(25+20k-4k^2)-84-120k=0

100+80k+16k^2-84-120k=0

16k^2-40k+16=0

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

8(2k-1) (k-2)=0

(16k-8) (k-2)=0

16k-8=0. Or k-2=0

16k=8. Or k=2

K=8/16

K=1/2. Or. K=2