Q. Solve the equation \(x^2 – x + 5 = 5\).

Answer

Solve:
\[
x^2 – x + 5 = 5
\]
Subtract 5 from both sides:
\[
x^2 – x = 0
\]
Factor:
\[
x(x-1)=0
\]
So
\[
x=0 \quad\text{or}\quad x=1.
\]

Final result: x = 0 or x = 1.

Detailed Explanation

Solve the equation: \(x^2 – x + 5 = 5\)

Subtract 5 from both sides to isolate the polynomial on one side:

\(x^2 – x + 5 – 5 = 5 – 5\)

Simplify:

\(x^2 – x = 0\)
Factor the left-hand side by factoring out the common factor \(x\):

\(x^2 – x = x(x – 1)\)
Use the zero-product property: if a product equals zero, at least one factor is zero. So set each factor equal to zero separately:

\(x = 0\) or \(x – 1 = 0\)

From \(x – 1 = 0\) we get \(x = 1\).
Check the solutions in the original equation:

For \(x = 0\): \(0^2 – 0 + 5 = 5\), true.

For \(x = 1\): \(1^2 – 1 + 5 = 5\), true.

Therefore, the solutions are \(x = 0\) and \(x = 1\).

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FAQs

How do you solve \(x^2 - x + 5 = 5\)?

-A: Subtract 5: \(x^2 - x = 0\). Factor: \(x(x-1)=0\). So \(x=0\) or \(x=1\).

How did you factor \(x^2 - x\)?

-A: Factor out \(x\): \(x^2-x=x(x-1)\). Set each factor to zero: \(x=0\) or \(x-1=0\).

Can I use the quadratic formula?

-A: Yes. For \(x^2-x=0\) use \(a=1,b=-1,c=0\): \(x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{1\pm1}{2}\), giving \(0\) and \(1\).

What is the discriminant and what does it tell me?

-A: Discriminant \(\Delta=b^2-4ac=(-1)^2-4(1)(0)=1>0\). That means two distinct real roots.

How can I check my solutions?

-A: Substitute \(x=0\) and \(x=1\) into the original \(x^2-x+5=5\). Both give \(0-0+5=5\) and \(1-1+5=5\), so they work.

What is the graphical interpretation?

-A: Graph \(y=x^2-x+5\) and the line \(y=5\). Their intersection x-values are \(0\) and \(1\). Equivalently, graph \(y=x^2-x\) and find zeros.

How would completing the square work here?

-A: From \(x^2-x=0\), write \((x-\tfrac12)^2-\tfrac14=0\). So \((x-\tfrac12)^2=\tfrac14\), giving \(x=\tfrac12\pm\tfrac12\), i.e. \(0\) and \(1\).

What if the equation was \(x^2-x+5=0\) instead?

-A: Then \(\Delta=1-20=-19<0\), so there are two complex conjugate roots: \(x=\frac{1\pm i\sqrt{19}}{2}\).

-Are there any extraneous solutions from the steps used?

-A: No. We only performed subtraction and factoring (reversible operations), so both found roots are valid.

What are the sum and product of the roots?

-A: For \(x^2-x=0\) the roots \(0\) and \(1\) have sum \(1\) and product \(0\). This matches \(-b/a=1\) and \(c/a=0\).

Could there be repeated roots?

-A: No. Discriminant \(1\neq0\) so the two roots are distinct (multiplicity 1 each).

Common mistakes to avoid?

-A: Don’t forget to move all terms to one side (subtract 5). -Avoid sign errors when factoring. Check solutions by substitution.

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