Q. \(x^2 – 6x = -6x + 4\).

Q: How do you solve x^2 - 6x = -6x + 4?

Add 6x to both sides to get x^2 = 4. Factor x^2 - 4 = (x - 2)(x + 2) = 0, so x = 2 or x = -2.

Q: What happens when like terms appear on both sides?

They can cancel. Here -6x on both sides cancels, leaving x^2 = 4 . If cancellation yields 0 = nonzero, there is no solution; 0 = 0 means infinitely many.

Q: How do I factor x^2 - 4?

Recognize a difference of squares: x^2 - 4 = (x - 2)(x + 2). Set each factor to zero to find solutions.

Q: Can I use the quadratic formula on x^2 - 4 = 0?.

Yes. With a = 1, b = 0, c = -4, x = (-0 \pm \sqrt{0^2 - 4\cdot 1\cdot (-4)))/(2\cdot 1) = \pm 2.

Q: How do I check my solutions x = 2 and x = -2?

How do I check my solutions x = 2 and x = -2?

Q: What is the geometric interpretation?

Solve x^2 - 6x = -6x + 4 as intersections of y = x^2 - 6x (parabola) and y = -6x + 4 (line). They meet at x = 2 (y = -8) and x = -2 (y = 16).

Answer

Solve
\[
x^2-6x=-6x+4.
\]
Add \(6x\) to both sides:
\[
x^2=4.
\]
So
\[
x=\pm 2.
\]

(Check: for \(x=2\) and \(x=-2\) both sides equal, so both are solutions.)

Detailed Explanation

Solve the equation step-by-step

Write the original equation:
\[ x^2 – 6x = -6x + 4 \]
Eliminate the \(-6x\) terms by adding \(6x\) to both sides (this cancels the \(-6x\) on each side):
\[ x^2 – 6x + 6x = -6x + 4 + 6x \]
Simplify both sides:
\[ x^2 = 4 \]
Solve for \(x\) by taking square roots of both sides. Remember to include the plus/minus:
\[ x = \pm \sqrt{4} = \pm 2 \]
List the solutions: \(x = 2\) and \(x = -2\)
Verify each solution in the original equation:
For \(x = 2\):
Left side: \(2^2 – 6(2) = 4 – 12 = -8\)
Right side: \(-6(2) + 4 = -12 + 4 = -8\)
They match, so \(x = 2\) is valid.

For \(x = -2\):
Left side: \((-2)^2 – 6(-2) = 4 + 12 = 16\)
Right side: \(-6(-2) + 4 = 12 + 4 = 16\)
They match, so \(x = -2\) is valid.

Final answer: \(x = 2\) or \(x = -2\)

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FAQs

How do you solve \(x^2 - 6x = -6x + 4\)?

Add \(6x\) to both sides to get \(x^2 = 4\). Factor \(x^2 - 4 = (x - 2)(x + 2) = 0\), so \(x = 2\) or \(x = -2\).

Why can I add \(6x\) to both sides?

Because performing the same operation on both sides of an equation preserves equality, and the solution set...

What happens when like terms appear on both sides?

They can cancel. Here \( -6x \) on both sides cancels, leaving \( x^2 = 4 \). If cancellation yields \( 0 = \) nonzero, there is no solution; \( 0 = 0 \) means infinitely many.

How do I factor \(x^2 - 4\)?

Recognize a difference of squares: \(x^2 - 4 = (x - 2)(x + 2)\). Set each factor to zero to find solutions.

Can I use the quadratic formula on \(x^2 - 4 = 0\)?.

Yes. With \(a = 1\), \(b = 0\), \(c = -4\), \(x = (-0 \pm \sqrt{0^2 - 4\cdot 1\cdot (-4)))/(2\cdot 1) = \pm 2\).

How do I check my solutions \(x = 2\) and \(x = -2\)?

Could any algebraic steps introduce extraneous solutions?

Not here. Adding equal terms and factoring are reversible for polynomials; always substitute solutions back into the original to confirm.

What is the geometric interpretation?

Solve \(x^2 - 6x = -6x + 4\) as intersections of \(y = x^2 - 6x\) (parabola) and \(y = -6x + 4\) (line). They meet at \(x = 2\) (\(y = -8\)) and \(x = -2\) (\(y = 16\)).

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