To find the value of \( n \) such that \( x^2 - 11x + n \) is a perfect square trinomial, let's recall the form of a perfect square trinomial. A trinomial of the form \( x^2 + 2ax + a^2 \) is a perfect square trinomial and can be written as \( (x + a)^2 \).
In your case, the trinomial is \( x^2 - 11x + n \). Let's compare it with the general form of the perfect square trinomial:
\[ x^2 - 11x + n = (x - a)^2 \]
Expanding the right-hand side, we get:
\[ (x - a)^2 = x^2 - 2ax + a^2 \]
By comparing the coefficients, we see that:
\[ -2a = -11 \implies a = \frac{11}{2} \]
\[ a^2 = n \implies n = \left(\frac{11}{2}\right)^2 = \frac{121}{4} \]
So, the value of \( n \) that makes \( x^2 - 11x + n \) a perfect square trinomial is:
\[ n = \frac{121}{4} \]
2.
Find the value of n such that x2 – 11x + n is a perfect square trinomial.
121
1 answer