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Solve the following equations by reducing them to quadratic equation
X⁴+5X²–36=0?
X⁴+5X²–36=0?
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Solve the following equations by reducing them to quadratic equation X...
Reducing a Quartic Equation to a Quadratic Equation
To solve the quartic equation X⁴ + 5X² – 36 = 0, we need to reduce it to a quadratic equation by making a substitution. Let's follow these steps:
Step 1: Let Y = X²
By substituting Y = X², we get:
Y² + 5Y – 36 = 0
Step 2: Solve the Quadratic Equation
Now, we have a quadratic equation in terms of Y, which can be easily solved using the quadratic formula:
Y = (-5 ± √(5² + 4*1*36)) / 2*1
Y = (-5 ± √(25 + 144)) / 2
Y = (-5 ± √169) / 2
Y = (-5 ± 13) / 2
Therefore, we have two possible values for Y:
Y₁ = (-5 + 13) / 2 = 4
Y₂ = (-5 - 13) / 2 = -9
Step 3: Substitute Back Y = X²
Now, substitute Y back as X² in each case:
For Y₁ = 4: X² = 4
X = ±√4
X = ±2
For Y₂ = -9: X² = -9
This case has imaginary roots as the square root of a negative number is not real.
Step 4: Final Solution
The solutions to the quartic equation X⁴ + 5X² – 36 = 0 are:
X = ±2
Therefore, the roots of the given quartic equation can be reduced to the quadratic equation X² = 4, giving us X = ±2 as the real solutions.
To solve the quartic equation X⁴ + 5X² – 36 = 0, we need to reduce it to a quadratic equation by making a substitution. Let's follow these steps:
Step 1: Let Y = X²
By substituting Y = X², we get:
Y² + 5Y – 36 = 0
Step 2: Solve the Quadratic Equation
Now, we have a quadratic equation in terms of Y, which can be easily solved using the quadratic formula:
Y = (-5 ± √(5² + 4*1*36)) / 2*1
Y = (-5 ± √(25 + 144)) / 2
Y = (-5 ± √169) / 2
Y = (-5 ± 13) / 2
Therefore, we have two possible values for Y:
Y₁ = (-5 + 13) / 2 = 4
Y₂ = (-5 - 13) / 2 = -9
Step 3: Substitute Back Y = X²
Now, substitute Y back as X² in each case:
For Y₁ = 4: X² = 4
X = ±√4
X = ±2
For Y₂ = -9: X² = -9
This case has imaginary roots as the square root of a negative number is not real.
Step 4: Final Solution
The solutions to the quartic equation X⁴ + 5X² – 36 = 0 are:
X = ±2
Therefore, the roots of the given quartic equation can be reduced to the quadratic equation X² = 4, giving us X = ±2 as the real solutions.
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Solve the following equations by reducing them to quadratic equation X⁴+5X²–36=0?
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Solve the following equations by reducing them to quadratic equation X⁴+5X²–36=0? for Class 10 2024 is part of Class 10 preparation. The Question and answers have been prepared according to the Class 10 exam syllabus. Information about Solve the following equations by reducing them to quadratic equation X⁴+5X²–36=0? covers all topics & solutions for Class 10 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Solve the following equations by reducing them to quadratic equation X⁴+5X²–36=0?.
Solve the following equations by reducing them to quadratic equation X⁴+5X²–36=0? for Class 10 2024 is part of Class 10 preparation. The Question and answers have been prepared according to the Class 10 exam syllabus. Information about Solve the following equations by reducing them to quadratic equation X⁴+5X²–36=0? covers all topics & solutions for Class 10 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Solve the following equations by reducing them to quadratic equation X⁴+5X²–36=0?.
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