given that 2x^2-kx+18 is a perfect square,find k and hence solve the equation 2x^2-kx+18=0
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
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
I just want to make it clear that for perfect squares b^2=4ac.this is easier if you are familiar with quadratic equations
Especially how were you able to figure out that you would put 2 up there in place of k
To determine the value of k and solve the equation 2x^2 - kx + 18 = 0, we first need to determine if the quadratic trinomial is a perfect square. For a quadratic trinomial to be a perfect square, it must be in the form (x + a)^2, where a is a constant.
To check if the given quadratic trinomial can be expressed as a perfect square, we compare it to the general form (x + a)^2 = x^2 + 2ax + a^2.
By comparing the coefficients of the given trinomial and the general form, we have:
2x^2 - kx + 18 = x^2 + 2ax + a^2
Since the coefficient of x^2 is already equal, we only need to focus on the coefficients of x and the constant term.
Comparing the coefficients of x:
- kx = 2ax
- k = -2a
Now, comparing the constant terms:
18 = a^2
Therefore, we have two equations to solve simultaneously:
1) k = -2a
2) a^2 = 18
From equation 2) we can find the possible values for a. Taking the square root of both sides:
√(a^2) = √18
a = ±√18
a = ±3√2
Substituting the value of a in equation 1) to find k:
k = -2(a)
k = -2(±3√2)
k = ±6√2 (- because -2(±3√2) is equal to ±6√2)
Therefore, the possible values for k are ±6√2.
To solve the equation 2x^2 - kx + 18 = 0 using the determined values of k, substitute them back into the equation and solve for x:
When k = 6√2:
2x^2 - 6√2x + 18 = 0
When k = -6√2:
2x^2 + 6√2x + 18 = 0
You can solve these equations by factoring, completing the square, or the quadratic formula to find the precise values of x.