Solve the Quadratic Equation for X: ax + a² = (a + x)² + 3

Q: Why does this equation have no real solutions?

The discriminant $ \Delta = a^2 - 12 $ is negative when \( a . When the discriminant is negative, the quadratic equation has no real solutions - only complex ones!

Q: What if a = 3 or a > 3?

If $ a \geq \sqrt{12} \approx 3.46 $, then $ a^2 \geq 12 $ and the discriminant becomes non-negative, giving real solutions. But the problem specifies \( a .

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

Always check the discriminant $ \Delta = b^2 - 4ac $ for any quadratic equation $ ax^2 + bx + c = 0 $. It tells you if real solutions exist before you try to solve!

Q: Can I still use the quadratic formula here?

You could use the quadratic formula, but it would give complex solutions involving $ \sqrt{negative} $. Since we want real solutions and \( a , the answer is simply "no solution".

Q: Did I make an algebra mistake if I got no solution?

No! Sometimes equations genuinely have no real solutions. Always trust your algebra - if the discriminant is negative under the given conditions, "no solution" is the correct answer.

Quadratic Equations with No Real Solutions

Look at the equation below and express X in terms of a, given that a<3.

$ax+a^2=(a+x)^2+3$

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations

00:00 Express X in terms of A

00:04 Expand brackets properly according to shortened multiplication formulas

00:15 Arrange the equation so that one side equals 0

00:23 Arrange the equation

00:26 Look at the coefficients (every number is actually multiplied by 1)

00:32 Use the quadratic formula to find possible solutions

00:43 Substitute appropriate values and solve to find the solutions

01:06 A is less than 3 according to the given data

01:09 Therefore, A squared is definitely less than 9

01:13 Thus, the above root expression is not logical - less than 0

01:19 Therefore, there is no solution

Step-by-step written solution

Follow each step carefully to understand the complete solution

Understand the problem

Look at the equation below and express X in terms of a, given that a<3.

$ax+a^2=(a+x)^2+3$

Step-by-step solution

To find $x$ in terms of $a$ from the equation $ax + a^2 = (a + x)^2 + 3$ , follow these steps:

Step 1: Expand the right-hand side:
$(a + x)^2 = a^2 + 2ax + x^2$
Step 2: Substitute and simplify the equation:
$ax + a^2 = a^2 + 2ax + x^2 + 3$
Step 3: Rearrange the terms:
$ax + a^2 - a^2 = 2ax + x^2 + 3$
$ax = 2ax + x^2 + 3$
Step 4: Move all terms to one side of the equation to form a quadratic:
$0 = x^2 + ax + 3$
Step 5: Analyze the quadratic equation $x^2 + ax + 3 = 0$ :
Calculate the discriminant: $\Delta = a^2 - 4 \cdot 1 \cdot 3 = a^2 - 12$ .
Step 6: Consider the condition for real solutions ( $\Delta \geq 0$ ):
Since $a^2 - 12 < 0$ for $a < 3$ , the discriminant is negative for values of $a < 3$ . This means the quadratic equation has no real solutions.

Therefore, for the given condition $a < 3$ , there is no solution for $x$ .

Final Answer

No solution

Key Points to Remember

Essential concepts to master this topic

Expansion Rule: Expand $(a+x)^2 = a^2 + 2ax + x^2$ completely
Rearrangement: Move all terms to one side: $x^2 + ax + 3 = 0$
Discriminant Check: Calculate $\Delta = a^2 - 12 < 0$ when $a < 3$ ✓

Common Mistakes

Avoid these frequent errors

Forgetting to expand the squared binomial completely
Don't just expand $(a+x)^2$ as $a^2 + x^2$ = missing the middle term! This gives a wrong equation and incorrect solutions. Always remember the middle term: $(a+x)^2 = a^2 + 2ax + x^2$ .

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

Why does this equation have no real solutions?

The discriminant $\Delta = a^2 - 12$ is negative when $a < 3$ . When the discriminant is negative, the quadratic equation has no real solutions - only complex ones!

What if a = 3 or a > 3?

If $a \geq \sqrt{12} \approx 3.46$ , then $a^2 \geq 12$ and the discriminant becomes non-negative, giving real solutions. But the problem specifies $a < 3$ .

How do I know when to check the discriminant?

Always check the discriminant $\Delta = b^2 - 4ac$ for any quadratic equation $ax^2 + bx + c = 0$ . It tells you if real solutions exist before you try to solve!

Can I still use the quadratic formula here?

You could use the quadratic formula, but it would give complex solutions involving $\sqrt{negative}$ . Since we want real solutions and $a < 3$ , the answer is simply "no solution".

Did I make an algebra mistake if I got no solution?

No! Sometimes equations genuinely have no real solutions. Always trust your algebra - if the discriminant is negative under the given conditions, "no solution" is the correct answer.

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