what is the value of c such that x^2-11x+c is a perfect-square trinomial?
You can find what c is by completing the square.
y = x² - 11x + \(\frac{121}{4}\) and this simplifies to (x - \(\frac{11}{2}\))²
So, c is \(\frac{121}{4}\)
Answer:
The value of c is \(\frac{121}{4}\)
Step-by-step explanation:
A perfect square trinomial is trinomial that can be written as the square of a binomial.
If, \(x^2-11x+c\) is a perfect square trinomial,
Then, we can write,
\(x^2-11x+c=(x+a)^2\) ---------(1)
Where a is any real number,
\(x^2-11x+c=x^2+2ax+a^2\)-------(2)
By comparing the coefficient of x,
We get,
\(2ax = -11\)
\(\implies a = -\frac{11}{2}\)
By substitution the value of a in equation (2),
\(x^2-11x+c=x^2+2\times -\frac{11}{2}x+(-\frac{11}{2})^2\)
\(\implies x^2-11x+c=x^2-11x+\frac{121}{4}\)
Again by comparing,
\(c=\frac{121}{4}\)
Hence, The value of c is \(\frac{121}{4}\)