The roots of the equation x4 + x3 - 4x2 + x + 1 = 0 are diminished by h so that the transformed equation does not contain x2 term. If the values of such h are α and β, then 12(α - β)2 =
35
25
105
115
To solve this problem, we are given a quartic equation: \( x^4 + x^3 - 4x^2 + x + 1 = 0 \), and we are asked to find \( 12(\alpha - \beta)^2 \) when the roots are diminished by a constant \( h \) so that the transformed equation no longer contains the \( x^2 \) term.
1. Understanding the Polynomial Equation:
The given equation is \( x^4 + x^3 - 4x^2 + x + 1 = 0 \), where the roots of the equation are diminished by \( h \) so that the transformed equation no longer has the \( x^2 \) term. This means that the new equation will not have the coefficient of \( x^2 \) after shifting the roots of the original equation by \( h \).
2. Applying Vieta's Formulas:
We use Vieta's formulas to relate the sum and product of the roots of the original equation to the coefficients. The sum of the roots of the equation is given by the negative of the coefficient of \( x^3 \), which is -1. For the transformed equation, we adjust the roots by \( h \), and the new sum of the roots will be influenced by the value of \( h \).
3. Solving for \( \alpha \) and \( \beta \):
Through the calculations, the transformed equation has roots \( \alpha \) and \( \beta \), which are related to the values of the roots of the original equation. We are asked to find \( 12(\alpha - \beta)^2 \), which involves manipulating the coefficients and roots of the transformed equation.
4. Final Calculation:
From the given options and based on the transformations, we determine that \( 12(\alpha - \beta)^2 = 35 \).
Final Answer:
The value of \( 12(\alpha - \beta)^2 \) is \( 35 \).
Let \( \alpha, \beta \) be the roots of the equation \( x^2 - ax - b = 0 \) with \( \text{Im}(\alpha) < \text{Im}(\beta) \). Let \( P_n = \alpha^n - \beta^n \). If \[ P_3 = -5\sqrt{7}, \quad P_4 = -3\sqrt{7}, \quad P_5 = 11\sqrt{7}, \quad P_6 = 45\sqrt{7}, \] then \( |\alpha^4 + \beta^4| \) is equal to:
A polynomial that has two roots or is of degree 2 is called a quadratic equation. The general form of a quadratic equation is y=ax²+bx+c. Here a≠0, b, and c are the real numbers.
Consider the following equation ax²+bx+c=0, where a≠0 and a, b, and c are real coefficients.
The solution of a quadratic equation can be found using the formula, x=((-b±√(b²-4ac))/2a)
Read More: Nature of Roots of Quadratic Equation