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

Q: How do you solve x^2 - x + 5 = 5?

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

Q: How did you factor x^2 - x?

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

Q: Can I use the quadratic formula?

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.

Q: What is the discriminant and what does it tell me?

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

Q: How can I check my solutions?

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.

Q: How would completing the square work here?

From x^2-x=0, write (x-\frac{1}{2})^2-\frac{1}{4}=0. So (x-\frac{1}{2})^2=\frac{1}{4}, giving x=\frac{1}{2} \pm \frac{1}{2}, i.e., 0 and 1.

Q: What if the equation was x^2-x+5=0 instead?

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

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 \(x = 1\).

See full solution

Need help? Try our AI homework tools: instant, accurate math help.

Get Help

FAQs

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

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

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

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?

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?

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

How can I check my solutions?

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?

How would completing the square work here?

From \(x^2-x=0\), write \((x-\frac{1}{2})^2-\frac{1}{4}=0\). So \((x-\frac{1}{2})^2=\frac{1}{4}\), giving \(x=\frac{1}{2} \pm \frac{1}{2}\), i.e., 0 and 1.

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

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

Math AI tools solve different problems.
Find your favorite today!

173,935+ happy customers
Math, Calculus, Geometry, etc.

✅ AI math helper 👤 AI geometry solver ✏️ AI calculus help