Solve the Equation: -2x(3-x)+(x-3)² = 9 for Parameter x

Q: Why do I need to expand everything first?

Expanding helps you combine like terms and see the equation's true form. Without expanding $ -2x(3-x) $ and $ (x-3)^2 $, you can't simplify to get the standard quadratic form.

Q: How do I remember the formula for $ (x-3)^2 $ ?

Use the pattern $ (a-b)^2 = a^2 - 2ab + b^2 $. So $ (x-3)^2 = x^2 - 2(x)(3) + 3^2 = x^2 - 6x + 9 $. The middle term is always twice the product!

Q: Why does $ 3x(x-4) = 0 $ give me two answers?

This uses the Zero Product Property: if two factors multiply to zero, at least one must be zero. So either $ 3x = 0 $ (giving x = 0) or $ x-4 = 0 $ (giving x = 4).

Q: Can I solve this without factoring?

Yes! You could use the quadratic formula on $ 3x^2 - 12x = 0 $, but factoring is much faster here since there's no constant term.

Q: How do I check if both answers are correct?

Substitute each value back into the original equation. For x = 0: $ -2(0)(3) + (0-3)^2 = 0 + 9 = 9 $ ✓. For x = 4: $ -2(4)(-1) + (4-3)^2 = 8 + 1 = 9 $ ✓

Quadratic Equations with Factoring Methods

Find the value of the parameter x.

$-2x(3-x)+(x-3)^2=9$

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations

00:09 Let's solve this problem step by step.

00:16 First, open the parentheses and multiply each factor inside.

00:26 Use the multiplication formulas to expand the parentheses.

00:39 Now, calculate the products from these multiplications.

00:47 Reduce by simplifying where possible.

00:52 Next, group similar terms together.

01:03 Break down each multiplication by using the factor three.

01:12 Take out the common factor from the entire expression.

01:25 Determine when each factor equals zero.

01:29 Then, isolate the unknown variable to find a solution.

01:34 That's one solution. Now, let's use the same method to find another one.

01:39 And there you have it. This is how we solve the problem!

Step-by-step written solution

Follow each step carefully to understand the complete solution

Understand the problem

Find the value of the parameter x.

$-2x(3-x)+(x-3)^2=9$

Step-by-step solution

To solve the equation $-2x(3-x) + (x-3)^2 = 9$ , follow these steps:

Step 1: Expand each term:
$-2x(3-x) = -6x + 2x^2$ and $(x-3)^2 = x^2 - 6x + 9$ .
Step 2: Substitute the expanded terms into the equation:
$-6x + 2x^2 + x^2 - 6x + 9 = 9$ .
Step 3: Combine like terms:
$(2x^2 + x^2) + (-6x - 6x) + 9 = 9$ .
Step 4: Simplify further:
$3x^2 - 12x + 9 = 9$ .
Step 5: Move all terms to one side to form a quadratic equation:
$3x^2 - 12x + 9 - 9 = 0$ .
Step 6: Simplify the expression:
$3x^2 - 12x = 0$ .
Step 7: Factor out the common term:
$3x(x - 4) = 0$ .
Step 8: Solve for $x$ :
Since $3x = 0$ or $x - 4 = 0$ , we find $x = 0$ or $x = 4$ .

Therefore, the values of $x$ that satisfy the equation are $\mathbf{x = 0}$ and $\mathbf{x = 4}$ .

Final Answer

$x=0,x=4$

Key Points to Remember

Essential concepts to master this topic

Expansion: Multiply terms carefully: $-2x(3-x) = -6x + 2x^2$
Factoring: Extract common factor: $3x^2 - 12x = 3x(x-4)$
Check: Substitute x = 0: $-2(0)(3) + (0-3)^2 = 9$ ✓

Common Mistakes

Avoid these frequent errors

Incorrectly expanding the squared term
Don't expand $(x-3)^2$ as $x^2 - 9$ = missing the middle term! This gives $3x^2 - 6x = 0$ instead of $3x^2 - 12x = 0$ . Always use $(a-b)^2 = a^2 - 2ab + b^2$ for complete expansion.

Practice Quiz

Test your knowledge with interactive questions

Find the value of the parameter x.

$$ (x-5)^2=0 $$

FAQ

Everything you need to know about this question

Why do I need to expand everything first?

Expanding helps you combine like terms and see the equation's true form. Without expanding $-2x(3-x)$ and $(x-3)^2$ , you can't simplify to get the standard quadratic form.

How do I remember the formula for $(x-3)^2$ ?

Use the pattern $(a-b)^2 = a^2 - 2ab + b^2$ . So $(x-3)^2 = x^2 - 2(x)(3) + 3^2 = x^2 - 6x + 9$ . The middle term is always twice the product!

Why does $3x(x-4) = 0$ give me two answers?

This uses the Zero Product Property: if two factors multiply to zero, at least one must be zero. So either $3x = 0$ (giving x = 0) or $x-4 = 0$ (giving x = 4).

Can I solve this without factoring?

Yes! You could use the quadratic formula on $3x^2 - 12x = 0$ , but factoring is much faster here since there's no constant term.

How do I check if both answers are correct?

Substitute each value back into the original equation. For x = 0: $-2(0)(3) + (0-3)^2 = 0 + 9 = 9$ ✓. For x = 4: $-2(4)(-1) + (4-3)^2 = 8 + 1 = 9$ ✓

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Solve the Equation: -2x(3-x)+(x-3)² = 9 for Parameter x

Quadratic Equations with Factoring Methods

❤️ Continue Your Math Journey!

Step-by-step video solution

Step-by-step written solution

Understand the problem

Step-by-step solution

Final Answer

Key Points to Remember

Common Mistakes

Practice Quiz

FAQ

Why do I need to expand everything first?

How do I remember the formula for $(x-3)^2$ ?

Why does $3x(x-4) = 0$ give me two answers?

Can I solve this without factoring?

How do I check if both answers are correct?

🌟 Unlock Your Math Potential

More Questions

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Solve (x-4)² + x(x-12) = 16: Finding Parameter x

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Right Triangle Side Lengths: Solve for x+9, x+2, and x+10

Solve the Equation: -2x(3-x)+(x-3)² = 9 for Parameter x

Quadratic Equations with Factoring Methods

❤️ Continue Your Math Journey!

Step-by-step video solution

Step-by-step written solution

Understand the problem

Step-by-step solution

Final Answer

Key Points to Remember

Common Mistakes

Practice Quiz

FAQ

Why do I need to expand everything first?

How do I remember the formula for (x−3)2 (x-3)^2 (x−3)2?

Why does 3x(x−4)=0 3x(x-4) = 0 3x(x−4)=0 give me two answers?

Can I solve this without factoring?

How do I check if both answers are correct?

🌟 Unlock Your Math Potential

More Questions

Solving Equations by Factoring

Solve (x-4)² + x(x-12) = 16: Finding Parameter x

Solve the Cubic Equation: 12x³-9x²-3x = 0

Right Triangle Side Lengths: Solve for x, x+2, and x+4

Right Triangle Side Lengths: Solve for x+9, x+2, and x+10

How do I remember the formula for $(x-3)^2$ ?

Why does $3x(x-4) = 0$ give me two answers?