Chapter 6: Problem 2
If \(a, b, c, d \in R\) then the equation \(\left(x^{2}+a x-3 b\right)\left(x^{2}-c x+b\right)\left(x^{2}-d x+2 b\right)=0\) has (1) 6 real roots (2) At least 2 real roots (3) 4 real roots (4) 3 real roots
Short Answer
Expert verified
The equation has at least 2 real roots.
Step by step solution
01
Identify the Factors
The given equation is \((x^2 + ax - 3b)(x^2 - cx + b)(x^2 - dx + 2b) = 0\). This equation is factored into three quadratic expressions. Each quadratic can potentially have two roots.
02
Roots of the First Quadratic
Consider the first quadratic \(x^2 + ax - 3b = 0\). The roots can be found using the quadratic formula: \(x = \frac{-a \pm \sqrt{a^2 + 12b}}{2}\). For real roots, the discriminant \(a^2 + 12b\) must be non-negative. To ensure at least some roots, assume \(b eq 0\).
03
Roots of the Second Quadratic
Consider the second quadratic \(x^2 - cx + b = 0\). The roots are given by: \(x = \frac{c \pm \sqrt{c^2 - 4b}}{2}\). For real roots, the discriminant \(c^2 - 4b\) must be non-negative. Again, assume \(b eq 0\).
04
Roots of the Third Quadratic
Consider the third quadratic \(x^2 - dx + 2b = 0\). The roots can be found using: \(x = \frac{d \pm \sqrt{d^2 - 8b}}{2}\). For real roots, the discriminant \(d^2 - 8b\) must be non-negative. Assume \(b eq 0\).
05
Conclusion on the Number of Real Roots
Each quadratic equation can have 0, 1, or 2 real roots based on its discriminant. Since each of the three quadratics can independently have at least one real root assuming \(b eq 0\), there are situations where all can have real roots. At a minimum, each quadratic will have at least one real root, resulting in at least two overall real roots among the three equations.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Quadratic Discriminant
To start with, let's understand the quadratic discriminant. In any quadratic equation of the form \ ax^2 + bx + c = 0 \, the discriminant is given by the expression: \ b^2 - 4ac \. This discriminant helps determine the nature of the roots:
- If \(b^2 - 4ac > 0\), the equation has two distinct real roots.
- If \(b^2 - 4ac = 0\), the equation has exactly one real root, which is also called a repeated or double root.
- If \(b^2 - 4ac < 0\), the equation has no real roots, instead it has two complex conjugate roots.
Real Roots
Real roots are solutions to a quadratic equation where the value of \(x\) is a real number. From earlier, we know that the discriminant's value determines whether the roots are real. Each quadratic in the given equation must have its discriminant tested to see if it results in real roots.
Consider the first quadratic equation \(x^2 + ax - 3b = 0\):
Consider the first quadratic equation \(x^2 + ax - 3b = 0\):
- The discriminant is \(a^2 + 12b\).
- If \(a^2 + 12b \geq 0\), the quadratic will have real roots.
- The discriminant is \(c^2 - 4b\).
- If \(c^2 - 4b \geq 0\), the quadratic will have real roots.
- The discriminant is \(d^2 - 8b\).
- If \(d^2 - 8b \geq 0\), the quadratic will have real roots.
Quadratic Formula
The quadratic formula provides the roots of any quadratic equation \(ax^2 + bx + c = 0\). The formula is given by: \ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \ When you apply the quadratic formula to each of the equations in the given problem, you can find their respective roots:
- For \(x^2 + ax - 3b = 0\), the roots are given by: \ x = \frac{-a \pm \sqrt{a^2 + 12b}}{2} \
- For \(x^2 - cx + b = 0\), the roots are given by: \ x = \frac{c \pm \sqrt{c^2 - 4b}}{2} \
- For \(x^2 - dx + 2b = 0\), the roots are given by: \ x = \frac{d \pm \sqrt{d^2 - 8b}}{2} \