Factor Theorem Application for $2x^2 - 5x + k = 0$
The problem asks us to find the value of k in the given quadratic equation, $2x^2 - 5x + k = 0$, knowing that $(x - 1)$ is a factor of this polynomial. We can solve this using the Factor Theorem.
Understanding the Factor Theorem
The Factor Theorem states that if $(x - a)$ is a factor of a polynomial $P(x)$, then $P(a) = 0$. In simpler terms, if you plug the root associated with the factor into the polynomial, the result will be zero.
Applying the Factor Theorem
In this case, our polynomial is $P(x) = 2x^2 - 5x + k$. The given factor is $(x - 1)$. According to the Factor Theorem, the root corresponding to this factor is found by setting $(x - 1) = 0$, which gives $x = 1$.
Since $(x - 1)$ is a factor, we must have $P(1) = 0$. Let's substitute $x = 1$ into the polynomial:
$P(1) = 2(1)^2 - 5(1) + k$
Calculating the Value of k
Now, we simplify the expression and set it equal to zero:
- $P(1) = 2(1) - 5(1) + k$
- $P(1) = 2 - 5 + k$
- $P(1) = -3 + k$
According to the Factor Theorem, $P(1)$ must equal 0. So, we set the expression equal to 0:
$-3 + k = 0$
To find the value of k, we solve this simple equation:
$k = 3$
Conclusion
Therefore, the value of k for which $(x - 1)$ is a factor of $2x^2 - 5x + k = 0$ is 3.
Important Questions from Algebra (Notes)
- What is the remainder when 2023²⁰²⁴ + 2025²⁰²⁴ is divided by 2024?
In an examination, a student scores 4 marks for every correct answer and loses 1 mark for every wrong answer. If she/he attempts all 60 questions and secures 130 marks, the number of questions she/he attempts wrongly, are?
Match List-I with List-II
List-1 List-II (A) If $\begin{bmatrix}\lambda-1 & 0 \\ 0 & \lambda-1 \end{bmatrix} $, then $\lambda$ is (I) 0 (B) If A=$ \begin{bmatrix}1 & 2 \\2 & 4 \end{bmatrix} $, then $\Delta$ is (II) 1 (C) If A = $ \begin{bmatrix}1 & 0 \\0 & \frac{1}{2} \end{bmatrix} $, then $|A^{-1}|$ is (III) -2 (D) If $ \begin{bmatrix}a+1 & 1 \\1 & 2 \end{bmatrix} = \begin{bmatrix}-1 & 1 \\1 & 2 \end{bmatrix} $, then a is (IV) 2 Choose the correct answer from the options given below:
- If $x = (2+\sqrt{3})^{\frac{1}{3}} + (2+\sqrt{3})^{-\frac{1}{3}}$ and $x^3-3x + k = 0$, then the value of k is:
If \[ \begin{vmatrix} \dfrac{1}{a(b+c)} & \dfrac{1}{b(c+a)} & \dfrac{1}{c(a+b)} \\ \dfrac{bc}{a(b+c)} & \dfrac{ca}{b(c+a)} & \dfrac{ab}{c(a+b)} \\ 1 & 1 & 1 \end{vmatrix} = k, \text{ then the value of } k \text{ is:} \]
