Solve the Quadratic Equation: 4x² + 1 = 0 with Complex Solutions

Quadratic Equations with Complex Solutions

Solve for x x :

4x2+1=0 4x^2+1=0

❤️ Continue Your Math Journey!

We have hundreds of course questions with personalized recommendations + Account 100% premium

Step-by-step video solution

Watch the teacher solve the problem with clear explanations
00:05 First, we need to find the value of X.
00:08 To do this, let's try to get X on its own.
00:20 Remember, any number squared is greater than zero, so it's always positive.
00:29 But here, our equation has X squared equal to a negative number.
00:39 So, there is no solution to this problem. And that's okay!

Step-by-step written solution

Follow each step carefully to understand the complete solution
1

Understand the problem

Solve for x x :

4x2+1=0 4x^2+1=0

2

Step-by-step solution

First, we should notice that it is a quadratic equation because there is a quadratic term (meaning raised to the second power).

The first step in solving a quadratic equation is always arranging it in a form where all terms on one side are ordered from highest to lowest power (in descending order from left to right) and 0 on the other side.

Then we can choose whether to solve it using the quadratic formula or by factoring/completing the square.

The equation in the problem is already arranged, so let's proceed with the solving technique:

We'll choose to solve it using the quadratic formula.

Let's recall it first:

The rule states that the roots of an equation in the formax2+bx+c=0 ax^2+bx+c=0 are x1,2=b±b24ac2a x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a} .

This formula is called: "The Quadratic Formula"

Let's now solve the problem:

4x2+1=0 4x^2+1=0

First, let's identify the coefficients of the terms:

{a=4b=0c=1 \begin{cases}a=4 \\ b=0 \\ c=1\end{cases}

Note that in the given equation there is no first-power term, so from comparing to the general form:

ax2+bx+c=0 ax^2+bx+c=0

we can conclude that the coefficient b b (which is the coefficient of the first-power term x x in the general form) is 0.

Let's continue and get the equation's solutions (roots) by substituting the coefficients we noted earlier in the quadratic formula:

x1,2=b±b24ac2a=0±0244124 x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{0\pm\sqrt{0^2-4\cdot4\cdot1}}{2\cdot4}

Let's continue and calculate the expression under the root and simplify the expression:

x1,2=±168 x_{1,2}=\frac{\pm\sqrt{-16}}{8}

We now have a negative expression under the root and since we cannot extract a real root from a negative number, this equation has no real solutions.

In other words, there is no real value of x x that when substituted in the equation will give a true statement.

Therefore, the correct answer is answer D.

3

Final Answer

No solution

Key Points to Remember

Essential concepts to master this topic
  • Discriminant: When b24ac<0 b^2 - 4ac < 0 , no real solutions exist
  • Technique: Apply quadratic formula: x=0±168 x = \frac{0 \pm \sqrt{-16}}{8}
  • Check: Negative under square root means no real solutions ✓

Common Mistakes

Avoid these frequent errors
  • Trying to find real solutions when discriminant is negative
    Don't attempt to simplify 16 \sqrt{-16} as a real number = impossible result! Real square roots of negative numbers don't exist. Always check the discriminant first: if b24ac<0 b^2 - 4ac < 0 , conclude no real solutions.

Practice Quiz

Test your knowledge with interactive questions

Solve the following equation:


\( 2x^2-8=x^2+4 \)

FAQ

Everything you need to know about this question

What does it mean when there are no real solutions?

+

It means there's no real number that makes the equation true. The graph of this parabola doesn't cross the x-axis, so there are no x-intercepts.

How do I know if a quadratic has real solutions before solving?

+

Calculate the discriminant b24ac b^2 - 4ac . If it's negative, no real solutions exist. If zero, one solution. If positive, two solutions.

Can I still use the quadratic formula when b = 0?

+

Absolutely! When there's no x term, b = 0. The formula still works: x=0±024ac2a x = \frac{0 \pm \sqrt{0^2 - 4ac}}{2a}

What are complex solutions?

+

Complex solutions involve imaginary numbers (using i=1 i = \sqrt{-1} ). For this problem, the complex solutions would be x=±i2 x = \pm \frac{i}{2} , but that's beyond basic algebra.

Why can't I just solve 4x² = -1?

+

You can rearrange it that way! You'd get x2=14 x^2 = -\frac{1}{4} . But since no real number squared gives a negative result, there's still no real solution.

🌟 Unlock Your Math Potential

Get unlimited access to all 18 Solving Quadratic Equations questions, detailed video solutions, and personalized progress tracking.

📹

Unlimited Video Solutions

Step-by-step explanations for every problem

📊

Progress Analytics

Track your mastery across all topics

🚫

Ad-Free Learning

Focus on math without distractions

No credit card required • Cancel anytime

More Questions

Click on any question to see the complete solution with step-by-step explanations