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Magazine Chapter 11: Problem 15
Solve. \(4 x^{4}+11 x^{2}=3\)
Short Answer
Expert verified
The real solutions are \( x = \frac{1}{2} \) and \( x = -\frac{1}{2} \).
Step by step solution
01
Substitute a New Variable
To simplify the equation, let's redefine it using substitution. Let us denote that \( u = x^2 \). This transforms our original equation into \( 4u^2 + 11u = 3 \).
02
Rearrange the Equation
Rearrange the equation to fit the standard quadratic form \( au^2 + bu + c = 0 \), which becomes \( 4u^2 + 11u - 3 = 0 \) after moving the 3 to the left side of the equation.
03
Apply the Quadratic Formula
Use the quadratic formula \( u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) to solve for \( u \). Here, \( a = 4 \), \( b = 11 \), and \( c = -3 \). Substitute these values to find the solutions for \( u \).
04
Calculate the Discriminant
Calculate the discriminant: \( 11^2 - 4\cdot 4 \cdot (-3) = 121 + 48 = 169 \). Since the discriminant is a perfect square, \( \sqrt{169} = 13 \), the equation has two real solutions.
05
Solve for Variable "u"
Calculate \( u \) using the quadratic formula: \( u = \frac{-11 \pm 13}{8} \). This gives two solutions: \( u = \frac{2}{8} = \frac{1}{4} \) and \( u = \frac{-24}{8} = -3 \).
06
Back-Substitute to Find x
Since \( u = x^2 \), solve for \( x \) in each case. For \( u = \frac{1}{4} \), \( x^2 = \frac{1}{4} \) so \( x = \pm \frac{1}{2} \). For \( u = -3 \), there is no real solution since \( x^2 = -3 \) yields no real number.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Variable Substitution
Variable substitution is a handy technique to simplify complex equations, especially when dealing with higher powers like x raised to the fourth power. Here, the original equation was transformed using a simple substitution: let \( u = x^2 \). This method turns the equation from a quartic (fourth-degree) equation into a quadratic, which is much easier to handle.
- Substitution is beneficial because it temporarily changes the form of an equation to something more manageable.
- Upon solving the simpler equation, you revert the substitution to find the original variable.
Quadratic Formula
The quadratic formula is a universal solution for any quadratic equation of the form \( au^2 + bu + c = 0 \). It is expressed as:\[u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]For our simplified equation \(4u^2 + 11u - 3 = 0\), this involved substituting \(a = 4\), \(b = 11\), and \(c = -3\).
- The quadratic formula helps find solutions for any quadratic equation as long as you know the coefficients.
- This approach is essential when factoring is difficult or impossible.
Discriminant
The discriminant is part of the quadratic formula, specifically the term under the square root: \(b^2 - 4ac\). It's crucial because it tells us about the nature of the roots or solutions a quadratic equation can have.
- If the discriminant is positive, there are two distinct real solutions.
- If it is zero, there is one real solution (a repeated root).
- If negative, no real solutions exist; the solutions are complex numbers.
Real Solutions
After calculating the possible values for \(u\) using the quadratic formula, we need to back-substitute and find solutions for \(x\).
- When \( u = \frac{1}{4} \), since \(u = x^2\), solving for \(x\) gives us \( x = \pm \frac{1}{2} \).
- When \( u = -3 \), the corresponding equation \(x^2 = -3\) has no real solution in the real number system, as square roots of negative numbers are complex.
