Solve each equation separately and find which x is the largest.$ -x^2+100=0 $$ 6x^2-44=5x^2-144 $

There is no solution to any equation

Solving x: Quadratics -x^2+100=0 and 6x²-44=5x²-144 Compared

Q: Why does the first equation have two solutions but the second has none?

The first equation gives us $ x^2 = 100 $ (positive), so x = ±10 both work. The second gives $ x^2 = -100 $ (negative), which is impossible for real numbers!

Q: How do I know which solution is larger when comparing equations?

First solve each equation completely. Then list all real solutions from both equations and pick the largest number. If one equation has no real solutions, ignore it completely.

Q: What does it mean when x² equals a negative number?

It means there are no real number solutions! You can't multiply any real number by itself to get a negative result. These are called imaginary solutions in advanced math.

Q: Do I always get two solutions from quadratic equations?

Not always! You get two solutions when x² = positive number, one solution when x² = 0, and no real solutions when x² = negative number.

Q: How do I avoid sign errors when moving terms around?

Write each step carefully! When moving $ -x^2 $ to the right side, it becomes $ +x^2 $. Always change the sign when moving terms across the equals sign.

Quadratic Equations with Real vs Imaginary Solutions

Solve each equation separately and find which x is the largest.

$-x^2+100=0$
$6x^2-44=5x^2-144$

❤️ 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

00:16 Let's find the largest value for X.

00:19 First, we'll look for X in section one.

00:29 We'll isolate X by moving terms around.

00:39 Taking the root helps us find possible solutions.

00:47 Here are the possible solutions from section one.

00:54 Next, let's calculate X from section two.

01:04 We'll move X to the left side of the equation.

01:33 Now, let's group similar terms together.

01:37 Remember, any number squared is at least zero.

01:41 Since minus 100 is less than zero, there's no solution in section two.

01:47 And that's how we solve this problem.

Step-by-step written solution

Follow each step carefully to understand the complete solution

Understand the problem

Solve each equation separately and find which x is the largest.

$-x^2+100=0$
$6x^2-44=5x^2-144$

Step-by-step solution

Let's solve each equation step-by-step:

1) Solve $-x^2 + 100 = 0$ :

Step 1: Isolate $x^2$ .

Rearrange the equation to $x^2 = 100$ .

Step 2: Solve for $x$ by taking the square root of both sides.

$x = \pm \sqrt{100}$
$x = \pm 10$

2) Solve $6x^2 - 44 = 5x^2 - 144$ :

Step 1: Simplify by moving all terms to one side.

$6x^2 - 5x^2 = -144 + 44$
$x^2 = -100$

Step 2: Since $x^2 = -100$ , there are no real solutions because we can't take the square root of a negative number in the set of real numbers.

Conclusion: From the solutions for our first equation, the possible values are $x = 10$ and $x = -10$ . The largest $x$ , since there are no real solutions from the second equation, is $x = 10$ .

Final Answer

2=1

Key Points to Remember

Essential concepts to master this topic

Rule: Square roots of negative numbers have no real solutions
Technique: Isolate x² first: -x² + 100 = 0 becomes x² = 100
Check: Substitute largest value back: -(10)² + 100 = 0 ✓

Common Mistakes

Avoid these frequent errors

Assuming all quadratic equations have real solutions
Don't assume every equation like x² = -100 has real answers = impossible square roots! Negative values under square roots have no real solutions. Always check if x² equals a positive number before finding 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

Why does the first equation have two solutions but the second has none?

The first equation gives us $x^2 = 100$ (positive), so x = ±10 both work. The second gives $x^2 = -100$ (negative), which is impossible for real numbers!

How do I know which solution is larger when comparing equations?

First solve each equation completely. Then list all real solutions from both equations and pick the largest number. If one equation has no real solutions, ignore it completely.

What does it mean when x² equals a negative number?

It means there are no real number solutions! You can't multiply any real number by itself to get a negative result. These are called imaginary solutions in advanced math.

Do I always get two solutions from quadratic equations?

Not always! You get two solutions when x² = positive number, one solution when x² = 0, and no real solutions when x² = negative number.

How do I avoid sign errors when moving terms around?

Write each step carefully! When moving $-x^2$ to the right side, it becomes $+x^2$ . Always change the sign when moving terms across the equals sign.

🌟 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