Question

plzz solve 6,7 and 8 hurry.....

Answer 1

6) 2x² - 8x + 7 = k
2x² - 8x + 7-k = 0
${b}^{2} - 4ac = 0$
b = -8
a = 2
c = 7-k

(-8)²- 4*2*(7-k) = 0
64 = 8*(7-k)
64 = 56 - 8k
64-56 = - 8k
8 = - 8k
k = -1
so, 2x²-8x+7-(-1) = 0
2x²-8x+8= 0
2x²-4x-4x+8=0
2x(x-2)-4(x-2)=0
(2x-4)(x-2)= 0

first root :
2x-4=0
2x=4
x=4/2
x=2

second root:
x-2=0
x= 2

therefore first and second roots are equal so k = -1

7) ax²+2bx+2=0
one root is twice than other root
b²-4ac = 2(b²-4ac)
b²-4ac = 2b² - 8ac
8ac - 4ac = 2b²-b²
b²= 4ac

8)x²+ax+b=0 and x²+bx+a=0 have common roots
so, x²+ax+b = x²+bx+a
x²-x²+ax-bx = a-b
x(a-b) = a-b
x = (a-b)/(a-b)
x= 1 substitute in any one of the two equations
so, 1²+b+a = 0
a+b+1=0
so a=b