The number of distinct real roots of the equation x5(x3 – x2 – x + 1) + x (3x3 – 4x2 – 2x + 4) – 1 = 0 is ______ .
The number of distinct real roots of the equation x5(x3 – x2 – x + 1) + x (3x3 – 4x2 – 2x + 4) – 1 = 0 is ______ .
Correct Answer: 3
Solution and Explanation
x8 – x7 – x6 + x5 + 3x4 – 4x3 – 2x2 + 4x – 1 = 0
⇒ x7(x – 1) – x5(x – 1) + 3x3(x – 1) – x (x2 – 1) + 2x (1 – x) + (x – 1) = 0
⇒ (x – 1) (x7 – x5 + 3x3 – x(x + 1) – 2x + 1) = 0
⇒ (x – 1) (x7 – x5 + 3x3 –x2 – 3x + 1) = 0
⇒ (x – 1) (x5 (x2 – 1) + 3x (x2 – 1) – 1 (x2 – 1)) = 0
⇒ (x – 1) (x2 – 1) (x5 + 3x – 1) = 0
∴ x = ± 1 are roots of above equation and x5 + 3x – 1 is a monotonic term hence vanishs at exactly one value of x other then 1 or – 1.
∴ 3 real roots.
So, the correct answer is 3.
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Concepts Used:
Quadratic Equations
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)
Two important points to keep in mind are:
- A polynomial equation has at least one root.
- A polynomial equation of degree ‘n’ has ‘n’ roots.
Read More: Nature of Roots of Quadratic Equation
There are basically four methods of solving quadratic equations. They are:
- Factoring
- Completing the square
- Using Quadratic Formula
- Taking the square root