Chapter 1: Problem 53
Find the solutions of the equation. $$4 x^{4}+25 x^{2}+36=0$$
Short Answer
Expert verified
There are no real solutions for the equation.
Step by step solution
01
Identify the Quadratic Structure
Notice that the equation \(4x^4 + 25x^2 + 36 = 0\) can be thought of as a quadratic in terms of \(x^2\). By letting \(y = x^2\), the equation becomes \(4y^2 + 25y + 36 = 0\).
02
Apply the Quadratic Formula
For the quadratic equation \(ay^2 + by + c = 0\), the solutions can be found using the quadratic formula: \[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]. Here, \(a = 4\), \(b = 25\), and \(c = 36\). Substitute these values into the formula.
03
Calculate the Discriminant
The discriminant \(b^2 - 4ac\) helps determine the nature of the roots. Calculate it as follows: \(b^2 - 4ac = 25^2 - 4 \times 4 \times 36 = 625 - 576 = 49\). Since the discriminant is positive, there will be two real roots.
04
Find the Roots for \(y\)
Substitute the discriminant and coefficients back into the quadratic formula: \[ y = \frac{-25 \pm 7}{8} \]. This provides two solutions: \(y = \frac{-25 + 7}{8} = -2.25\) and \(y = \frac{-25 - 7}{8} = -4\).
05
Solve for \(x\)
Recall that \(y = x^2\). Suppose \(x^2 = -2.25\) or \(x^2 = -4\). Since both are negative, there are no real solutions for \(x\) because the square roots of negative numbers are not real.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Quadratic Formula
When faced with a quadratic equation of the form \(ax^2 + bx + c = 0\), a powerful tool to find the solutions is the Quadratic Formula. This formula is expressed as:
\[y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
It allows you to solve for \(y\) by simply plugging in the coefficients \(a\), \(b\), and \(c\) from the quadratic equation. Here is a quick breakdown of how to use it:
\[y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
It allows you to solve for \(y\) by simply plugging in the coefficients \(a\), \(b\), and \(c\) from the quadratic equation. Here is a quick breakdown of how to use it:
- Identify the coefficients \(a\), \(b\), and \(c\) from your equation.
- Calculate the value of the discriminant \(b^2 - 4ac\).
- Substitute these values into the Quadratic Formula.
Discriminant
The discriminant, represented as \(b^2 - 4ac\), plays a crucial role in determining the nature of the roots for quadratic equations. Here's how it works:
- If the discriminant is positive, like in the example \(b^2 - 4ac = 49\), the quadratic equation will have two distinct real roots.
- If the discriminant equals zero, there is exactly one real root, which is also known as a repeated or double root.
- If the discriminant is negative, there are no real roots, but rather two complex roots.
Polynomial Equations
Polynomial equations involve terms that are non-negative integer powers of the variable, like \(x^4\) or \(x^2\). In the given exercise, the equation is initially in a quartic form:
\[4x^4 + 25x^2 + 36 = 0\]
To solve this kind of polynomial, identifying a pattern that allows it to resemble a more familiar form, like a quadratic equation, is essential. By substituting \(y = x^2\), the polynomial is transformed into a quadratic:
\[4y^2 + 25y + 36 = 0\]
This transformation is a powerful technique to simplify complex polynomial equations. With the resulting quadratic, standard techniques like the Quadratic Formula can be applied. Always remember:
\[4x^4 + 25x^2 + 36 = 0\]
To solve this kind of polynomial, identifying a pattern that allows it to resemble a more familiar form, like a quadratic equation, is essential. By substituting \(y = x^2\), the polynomial is transformed into a quadratic:
\[4y^2 + 25y + 36 = 0\]
This transformation is a powerful technique to simplify complex polynomial equations. With the resulting quadratic, standard techniques like the Quadratic Formula can be applied. Always remember:
- Transform complex polynomials into known formats.
- Use substitution effectively to make solving easier.
- Reinterpret the solutions back to the original variables if substitutions were made.