Given that 2x^2-kx+18 is a perfect square,find k and hence solve the equation 2x^2-kx+18=0 by factorisation.Plz explain in detail.
the 2nd order (squared) term must have a coefficient of one
dividing by 2 ... x^2 - k/2 x + 9 = 0
... take the square root of the constant term ... √9 = ± 3
factoring ... (x - 3)^2 = x^2 - 6 x + 9 ... k/2 = 6 ... k = 12
Why must it have a coefficient of one
To find the value of 'k' that makes the quadratic equation 2x^2 - kx + 18 a perfect square, we need to consider the general form of a perfect square trinomial.
A perfect square trinomial can be expressed as the square of a binomial. In general, it can be written as (ax + b)^2, where 'a' and 'b' are constants. Expanding this expression will yield a perfect square trinomial.
Let's try to write 2x^2 - kx + 18 as a perfect square trinomial by completing the square.
First, identify the coefficient of 'x^2', which is 2. This means 'a' in our perfect square expression should be √2x. We'll rewrite the equation as (√2x + b)^2.
Now, we will focus on obtaining the coefficient of 'x' in the square term. In our given equation, the coefficient of 'x' is -k. Therefore, when expanding (√2x + b)^2, we should have -kx.
Expanding (√2x + b)^2 will give us:
(√2x + b)^2 = (√2x)^2 + 2(√2x)(b) + b^2
= 2x + 2√2bx + b^2.
The coefficient of 'x' in the expansion is 2√2bx. We need this to be equal to -kx. Therefore, we have:
2√2b = -k.
Comparing coefficients of 'x' from the original equation and the perfect square expansion, we equate 2√2b to -k. We get the equation: 2√2b = -k.
Simplifying it further: b = -k/(2√2).
Now, let's compare the constant term from the original equation and the perfect square expansion. In our original equation, the constant term is 18. So, in the expansion (√2x + b)^2, we need the constant term to be b^2.
Therefore, we have: b^2 = 18.
Substituting the value of b = -k/(2√2), we can solve the equation for k:
(-k/(2√2))^2 = 18
k^2/(8) = 18
k^2 = 144
k = ±√144
k = ±12
Thus, the possible values for k are 12 and -12.
Now, let's solve the equation 2x^2 - kx + 18 = 0 by factorization using the values of k we found.
Case 1: k = 12
We have the equation 2x^2 - 12x + 18 = 0.
Factoring, we can divide the equation by 2 to simplify it:
x^2 - 6x + 9 = 0
This quadratic expression can be factored as:
(x - 3)(x - 3) = 0
(x - 3)^2 = 0
Taking the square root of both sides, we get:
x - 3 = 0
x = 3
Case 2: k = -12
We have the equation 2x^2 + 12x + 18 = 0.
Dividing by 2, we get:
x^2 + 6x + 9 = 0
This quadratic equation also factors as:
(x + 3)(x + 3) = 0
(x + 3)^2 = 0
Taking the square root of both sides, we get:
x + 3 = 0
x = -3
Therefore, the values of 'k' that make the given quadratic equation a perfect square are k = 12 and k = -12. The solutions to the equation 2x^2 - kx + 18 = 0 by factorization are x = 3 and x = -3.
must is probably to strong an imperative
... it makes the calculations less "messy"