Question:
How many real solutions are there for the following equation?
\[
x^4 + 5x^2 - 14 = 0
\]
How many real solutions are there for the following equation?
\[
x^4 + 5x^2 - 14 = 0
\]
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To solve quartic equations, use substitution and reduce them to quadratic equations.
Updated On: Sep 30, 2025
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The Correct Option is D
Solution and Explanation
Step 1: Introduce substitution.
Let \( y = x^2 \). The equation becomes: \[ y^2 + 5y - 14 = 0. \]
Step 2: Solve the quadratic equation.
We solve for \( y \) using the quadratic formula: \[ y = \frac{-5 \pm \sqrt{5^2 - 4(1)(-14)}}{2(1)} = \frac{-5 \pm \sqrt{25 + 56}}{2} = \frac{-5 \pm \sqrt{81}}{2} = \frac{-5 \pm 9}{2}.
\] Thus, \( y = 2 \) or \( y = -7 \).
Step 3: Solve for \( x \).
For \( y = 2 \), \( x^2 = 2 \Rightarrow x = \pm \sqrt{2} \). For \( y = -7 \), there are no real solutions since \( x^2 \) cannot be negative.
Step 4: Conclusion. Thus, there are 2 real solutions: \( x = \pm \sqrt{2} \).
Let \( y = x^2 \). The equation becomes: \[ y^2 + 5y - 14 = 0. \]
Step 2: Solve the quadratic equation.
We solve for \( y \) using the quadratic formula: \[ y = \frac{-5 \pm \sqrt{5^2 - 4(1)(-14)}}{2(1)} = \frac{-5 \pm \sqrt{25 + 56}}{2} = \frac{-5 \pm \sqrt{81}}{2} = \frac{-5 \pm 9}{2}.
\] Thus, \( y = 2 \) or \( y = -7 \).
Step 3: Solve for \( x \).
For \( y = 2 \), \( x^2 = 2 \Rightarrow x = \pm \sqrt{2} \). For \( y = -7 \), there are no real solutions since \( x^2 \) cannot be negative.
Step 4: Conclusion. Thus, there are 2 real solutions: \( x = \pm \sqrt{2} \).
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