Solve the equation x² − 5x + 6 = 0
\[ x^2-5x+6=0 \quad\Rightarrow\quad x=2 \text{ or } x=3 \]
Key idea: factor the quadratic. Find two numbers whose product is \(6\) and whose sum is \(-5\). Those are \(-2\) and \(-3\):
\[ x^2-5x+6=(x-2)(x-3) \]
So:
\[ (x-2)(x-3)=0 \]
By the zero-product principle, one factor must be zero:
\[ x-2=0 \quad\text{or}\quad x-3=0 \]
Hence:
\[ x=2 \quad\text{or}\quad x=3 \]
Check:
- For \(x=2\): \(2^2-5(2)+6=4-10+6=0\)
- For \(x=3\): \(3^2-5(3)+6=9-15+6=0\)
Both solutions work.
Обсуждение
4 сообщенияTry \(x^2-7x+12=0\) next.
\[ x^2-7x+12=0 \quad\Rightarrow\quad x=3 \text{ or } x=4 \]
Factor the quadratic: find two numbers with product \(12\) and sum \(-7\). They are \(-3\) and \(-4\):
\[ x^2-7x+12=(x-3)(x-4) \]
So:
\[ (x-3)(x-4)=0 \]
Thus:
\[ x-3=0 \quad\text{or}\quad x-4=0 \]
Hence:
\[ x=3 \quad\text{or}\quad x=4 \]
Check:
- \(x=3:\quad 9-21+12=0\)
- \(x=4:\quad 16-28+12=0\)
Both solutions are correct.
Try \(x^2-7x+12=0\) next.
That's the same equation as your previous message, so the answer is unchanged:
\[ x^2-7x+12=(x-3)(x-4)=0 \quad\Rightarrow\quad x=3 \text{ or } x=4 \]
If you meant a different equation, send the new one (or a photo of it) and I'll solve it.
Want a quick similar one to practice? Try \(x^2-9x+20=0\) — look for two numbers with product \(20\) and sum \(-9\).
