Solve the Quadratic Equation: x² + 5x + 10 = 0

Q: What does it mean when a quadratic has no solution?

It means the parabola never crosses the x-axis. Since $ x^2 + 5x + 10 = 0 $ opens upward and stays above the x-axis, there are no real x-values that make it equal zero.

Q: How do I know if there are no real solutions without graphing?

Calculate the discriminant $ \Delta = b^2 - 4ac $. If it's negative, there are no real solutions. For our equation: \( 25 - 40 = -15 , so no real solutions!

Q: Are there any solutions at all, or just no real ones?

There are complex solutions involving imaginary numbers, but in basic algebra we typically say "no solution" when we mean no real solutions.

Q: Can I still use the quadratic formula here?

You could, but you'd get $ x = \frac{-5 \pm \sqrt{-15}}{2} $, which involves $ \sqrt{-15} $. Since we can't take the square root of negative numbers in real numbers, it confirms: no real solutions.

Q: What if I made an error calculating the discriminant?

Double-check: $ a = 1, b = 5, c = 10 $, so $ \Delta = 5^2 - 4(1)(10) = 25 - 40 = -15 $. Always be careful with signs when substituting values!

Quadratic Equations with No Real Solutions

$x^2+5x+10=0$

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations

00:00 Find X

00:03 Pay attention to the coefficients

00:13 Use the roots formula

00:23 Substitute appropriate values according to the given data and solve for X

00:49 Calculate the products and the square

01:04 There's no such thing as a root of a negative number, therefore there is no solution

01:10 And this is the solution to the question

Step-by-step written solution

Follow each step carefully to understand the complete solution

Understand the problem

$x^2+5x+10=0$

Step-by-step solution

To solve this problem, we'll follow these steps:

Step 1: Calculate the discriminant of the quadratic equation.
Step 2: Use the discriminant to determine the number and type of solutions.

Step 1: Calculate the discriminant using the formula:

$\Delta = b^2 - 4ac = 5^2 - 4 \times 1 \times 10 = 25 - 40 = -15$ .

Step 2: Analyze the discriminant:

Since the discriminant ( $\Delta$ ) is negative ( $-15$ ), this indicates that the quadratic equation has no real solutions.

Therefore, the final solution is that the equation $x^2 + 5x + 10 = 0$ has no solution.

Comparing this with the given answer choices, the correct choice is:

Choice 3: No solution

Final Answer

No solution

Key Points to Remember

Essential concepts to master this topic

Rule: Calculate discriminant to determine solution types: $\Delta = b^2 - 4ac$
Technique: When $\Delta < 0$ , equation has no real solutions
Check: Verify $5^2 - 4(1)(10) = 25 - 40 = -15 < 0$ ✓

Common Mistakes

Avoid these frequent errors

Trying to use quadratic formula when discriminant is negative
Don't attempt to find real solutions when $\Delta = -15 < 0$ = imaginary answers only! The square root of a negative number isn't real. Always check the discriminant first to determine if real solutions exist.

Practice Quiz

Test your knowledge with interactive questions

a = Coefficient of x²

b = Coefficient of x

c = Coefficient of the independent number

what is the value of $$ a $$ in the equation

$$ y=3x-10+5x^2 $$

FAQ

Everything you need to know about this question

What does it mean when a quadratic has no solution?

It means the parabola never crosses the x-axis. Since $x^2 + 5x + 10 = 0$ opens upward and stays above the x-axis, there are no real x-values that make it equal zero.

How do I know if there are no real solutions without graphing?

Calculate the discriminant $\Delta = b^2 - 4ac$ . If it's negative, there are no real solutions. For our equation: $25 - 40 = -15 < 0$ , so no real solutions!

Are there any solutions at all, or just no real ones?

There are complex solutions involving imaginary numbers, but in basic algebra we typically say "no solution" when we mean no real solutions.

Can I still use the quadratic formula here?

You could, but you'd get $x = \frac{-5 \pm \sqrt{-15}}{2}$ , which involves $\sqrt{-15}$ . Since we can't take the square root of negative numbers in real numbers, it confirms: no real solutions.

What if I made an error calculating the discriminant?

Double-check: $a = 1, b = 5, c = 10$ , so $\Delta = 5^2 - 4(1)(10) = 25 - 40 = -15$ . Always be careful with signs when substituting values!

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