Q. \(x^2 – 6x = -6x + 4\)

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

What is the first step to solve \(x^2-6x=-6x+4\)?

-A: -Add \(6x\) to both sides to cancel the linear term: \(x^2=4\). Then solve \(x^2=4\) by square roots or factoring.

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

-A: By the equality property, performing the same operation on both sides preserves equality, so adding \(6x\) to each side is valid.

How do I get the solutions from \(x^2=4\)?

-A: Take square roots: \(x=\pm 2\). Or factor: \(x^2-4=(x-2)(x+2)=0\), giving \(x=2\) or \(x=-2\).

-Are both \(x=2\) and \(x=-2\) valid in the original equation?

-A: Yes. Substitute each into \(x^2-6x=-6x+4\); both satisfy the equation, so neither is extraneous.

Could I divide both sides by \(x\) to simplify?

-A: No-dividing by \(x\) risks losing the solution \(x=0\). -Always avoid dividing by a variable unless you know it’s nonzero.

What if the equation were \(x^2-6x=-6x\)?

-A: Then adding \(6x\) gives \(x^2=0\), so the only solution is \(x=0\) (a double root).

How to factor the quadratic form here?

-A: From \(x^2-4=0\), factor as \((x-2)(x+2)=0\). Set each factor to zero to find solutions.

How would I solve it by completing the square?

-A: Here it’s unnecessary because \(x^2=4\) is immediate. In general, write \(x^2-6x+\square = -6x+4+\square\) and proceed, but simpler methods apply.

What is the graphical interpretation?

-A: Solutions are intersection points of \(y=x^2\) and \(y=4\). Those intersections occur at \(x=\pm2\).

What common mistakes should I avoid?

-A: Don’t cancel identical terms incorrectly, don’t divide by a variable, and remember the ± when taking square roots.

Is this a quadratic equation and what is its degree?

-A: Yes, it’s quadratic (degree 2). -After simplification it becomes \(x^2=4\).

How many real solutions does it have?

-A: Two real solutions: \(x=2\) and \(x=-2\).

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