given that 2x^2-kx+18 is a perfect square,find k and hence solve the equation 2x^2-kx+18=0

Answer 1

2x^2-kx+18 = 0
Let's make that 2x^2-2nx+18 = 0
Now we can divide by 2, and we get
2(x^2-nx+9) = 0
Now, note that 9 = 3^2, so we have
x^2-nx+9 = (x-3)^2 = x^2-6x+9
Clearly, n = 6
So, our original equation now becomes
2x^2-12x+18 = 0
2(x-3)^2 = 0
x = 3

Answer 2

Sir obleck please explain to me how did you get to replace
2x²-2nx+18?

I want to learn it

Or you could say
a=2 b=-k c=18
For perfect square
B²=4ac

k²=4(2)(18)

K=√(144)=12

And you would still have

2x²-12x+18=0

Answer 3

Especially how were you able to figure out that you would put 2 up there in place of k

Answer 4

I just want to make it clear that for perfect squares b^2=4ac.this is easier if you are familiar with quadratic equations