Solve the Quadratic Equation: x²+3x-4=2x²

Q: What does 'no solution' really mean?

No real solution means the equation has no answers using regular numbers. The parabola $ y = x^2 - 3x + 4 $ never touches the x-axis, so there's no x-value that makes y = 0.

Q: How do I know when to check the discriminant?

Always calculate the discriminant $ b^2 - 4ac $ for any quadratic equation! If it's negative, stop there - no real solutions exist. If it's positive or zero, proceed with the quadratic formula.

Q: Can I still use the quadratic formula with a negative discriminant?

Technically yes, but you'll get complex numbers involving $ \sqrt{-7} $. For most algebra courses, we stop at 'no real solution' when the discriminant is negative.

Q: Why did we multiply by -1 in the solution?

Multiplying by -1 gives us positive leading coefficient $ x^2 - 3x + 4 = 0 $ instead of $ -x^2 + 3x - 4 = 0 $. Both forms are correct, but positive leading coefficients are easier to work with!

Q: What would the graph of this equation look like?

The parabola $ y = x^2 - 3x + 4 $ opens upward and sits entirely above the x-axis. Its vertex is at $ (1.5, 1.75) $, so it never crosses x-axis = no real roots.

Quadratic Equations with No Real Solutions

Solve the following equation:

$x^2+3x-4=2x^2$

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations

00:00 Find X

00:03 Arrange the equation so that one side equals 0

00:14 Group terms

00:24 Identify coefficients

00:34 Use the roots formula

00:51 Substitute appropriate values and solve

01:12 Calculate the square and products

01:34 It's not logical to have a root of a negative number

01:42 Therefore there is no solution to the question

Step-by-step written solution

Follow each step carefully to understand the complete solution

Understand the problem

Solve the following equation:

$x^2+3x-4=2x^2$

Step-by-step solution

Given the equation:

$x^2 + 3x - 4 = 2x^2$

Step 1: Move all terms to one side:

Subtract $2x^2$ from both sides to get:

$x^2 + 3x - 4 - 2x^2 = 0$

Simplify this to:

$-x^2 + 3x - 4 = 0$

Step 2: Rearrange to standard form:

Multiply the entire equation by -1 for simplicity:

$x^2 - 3x + 4 = 0$

Step 3: Solve the quadratic equation using the quadratic formula:

Here, $a = 1$ , $b = -3$ , $c = 4$ .

Plug into the quadratic formula:

$x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4 \times 1 \times 4}}{2 \times 1}$

$x = \frac{3 \pm \sqrt{9 - 16}}{2}$

$x = \frac{3 \pm \sqrt{-7}}{2}$

Step 4: Interpret the result:

The discriminant ( $b^2 - 4ac$ ) is negative, $-7$ , indicating no real solutions.

Conclusion: The equation has no solution in the set of real numbers.

In comparison with the provided choices, the correct choice is:

No solution

Final Answer

No solution

Key Points to Remember

Essential concepts to master this topic

Standard Form: Move all terms to one side to get $ax^2 + bx + c = 0$
Discriminant: Calculate $b^2 - 4ac$ to determine solution type: 9 - 16 = -7
Check: Negative discriminant means no real solutions exist ✓

Common Mistakes

Avoid these frequent errors

Assuming all quadratic equations have real solutions
Don't automatically try to find x-values when the discriminant is negative = wasted time and confusion! A negative discriminant means the parabola doesn't cross the x-axis. 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 'no solution' really mean?

No real solution means the equation has no answers using regular numbers. The parabola $y = x^2 - 3x + 4$ never touches the x-axis, so there's no x-value that makes y = 0.

How do I know when to check the discriminant?

Always calculate the discriminant $b^2 - 4ac$ for any quadratic equation! If it's negative, stop there - no real solutions exist. If it's positive or zero, proceed with the quadratic formula.

Can I still use the quadratic formula with a negative discriminant?

Technically yes, but you'll get complex numbers involving $\sqrt{-7}$ . For most algebra courses, we stop at 'no real solution' when the discriminant is negative.

Why did we multiply by -1 in the solution?

Multiplying by -1 gives us positive leading coefficient $x^2 - 3x + 4 = 0$ instead of $-x^2 + 3x - 4 = 0$ . Both forms are correct, but positive leading coefficients are easier to work with!

What would the graph of this equation look like?

The parabola $y = x^2 - 3x + 4$ opens upward and sits entirely above the x-axis. Its vertex is at $(1.5, 1.75)$ , so it never crosses x-axis = no real roots.

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