Question

x^2 - 2k(x+1) +k^2 = 6 has real roots. Find the roots in terms of k. Ans given: sqrt(2k+6) and - sqrt(2k+6)

Answer 1

is this a repeat?

Answer 2

Since the equation has real roots, then :
\[\Delta\geq0\]
The standard form of the equation :
\[x^2-2kx+k^2-2k-6=0\]
Then :
\[\Delta=4k^2-4\times(k^2-2k-6)=8k+24=8(k+3)\]
So the roots
\[x_1=\frac{2k+2\sqrt{2(k+3)}}{2}=k+\sqrt{2(k+3)}
\\x_2=k-\sqrt{2(k+3)}
\]

Answer 3

yeah i was totally wrong

Answer 4

@Noura11 I dont understand how you got to Δ=4k^2−4×(k^2−2k−6) sorry :/

Answer 5

OK !
The standard form of the equation :
\[x^2−2kx+k^2−2k−6=0\]
Then :
\[a=1~~ b=-2k~~~ c=k^2-2k+6\]
And we know :
\[\Delta= b^2-4\times a\times c\]