The problem requires us to find the value of $k$ in the cubic equation $x3 - 3x + k = 0$, given a specific expression for $x$.
Simplifying the Expression for x
We are given:
$x = (2+\sqrt{3})^{\frac{1}{3}} + (2+\sqrt{3})^{-\frac{1}{3}}$
Let's simplify this expression. Let:
$a = (2+\sqrt{3})^{\frac{1}{3}}$
And
$b = (2+\sqrt{3})^{-\frac{1}{3}}$
So, $x = a + b$.
Calculating x3
To relate $x$ to the given cubic equation, we can cube the expression for $x$:
$x3 = (a + b)3$
Using the algebraic identity $(a+b)3 = a3 + b3 + 3ab(a+b)$, we get:
$x3 = a3 + b3 + 3ab(x)$
Calculating a3 and b3
Now, let's calculate $a3$ and $b3$:
- $a3 = \left( (2+\sqrt{3})^{\frac{1}{3}} \right)3 = 2+\sqrt{3}$
- $b3 = \left( (2+\sqrt{3})^{-\frac{1}{3}} \right)3 = (2+\sqrt{3})^{-1}$
To simplify $b3$, we rationalize the denominator:
$b3 = \frac{1}{2+\sqrt{3}}$
Multiply the numerator and denominator by the conjugate of the denominator, which is $(2-\sqrt{3})$:
$b3 = \frac{1}{2+\sqrt{3}} \times \frac{2-\sqrt{3}}{2-\sqrt{3}} = \frac{2-\sqrt{3}}{2^2 - (\sqrt{3})^2} = \frac{2-\sqrt{3}}{4-3} = \frac{2-\sqrt{3}}{1} = 2-\sqrt{3}$
Calculating a3 + b3
$a3 + b3 = (2+\sqrt{3}) + (2-\sqrt{3}) = 2 + \sqrt{3} + 2 - \sqrt{3} = 4$
Calculating ab
$ab = (2+\sqrt{3})^{\frac{1}{3}} \times (2+\sqrt{3})^{-\frac{1}{3}}$
Using the exponent rule $ym * yn = ym+n$:
$ab = (2+\sqrt{3})^{\frac{1}{3} + (-\frac{1}{3})} = (2+\sqrt{3})^{\frac{1}{3} - \frac{1}{3}} = (2+\sqrt{3})^{0} = 1$
Substituting back into the x3 equation
Now substitute the values of $a3 + b3$ and $ab$ back into the equation for $x3$:
$x3 = (a3 + b3) + 3ab(x)$
$x3 = 4 + 3(1)(x)$
$x3 = 4 + 3x$
Finding the Value of k
Rearrange the equation to match the form $x3 - 3x + k = 0$:
$x3 - 3x - 4 = 0$
Now, compare this equation with the given equation:
$x3 - 3x + k = 0$
By direct comparison, we can see that:
$k = -4$
Final Answer Summary
The steps involved simplifying the expression for $x$, cubing it using the identity $(a+b)3$, calculating the components $a3$, $b3$, and $ab$, and substituting these back to get an equation relating $x3$ and $x$. Comparing this derived equation with the given equation $x3 - 3x + k = 0$ allowed us to determine the value of $k$.
The value of $k$ is -4.
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 - 1) is a factor of $2x^2 - 5x + k = 0$, then the value of k is:
If \[ \begin{vmatrix} \dfrac{1}{a(b+c)} & \dfrac{1}{b(c+a)} & \dfrac{1}{c(a+b)} \\ \dfrac{bc}{a(b+c)} & \dfrac{ca}{b(c+a)} & \dfrac{ab}{c(a+b)} \\ 1 & 1 & 1 \end{vmatrix} = k, \text{ then the value of } k \text{ is:} \]
