Solve the equation x² − 5x + 6 = 0
Answer: \(x=2\) or \(x=3\).
How to solve
\[ x^2-5x+6=0 \]
To factor, find two numbers that multiply to \(6\) and add to \(-5\). Those numbers are \(-2\) and \(-3\):
\[ x^2-5x+6=(x-2)(x-3) \]
So:
\[ (x-2)(x-3)=0 \]
A product is zero when one factor is zero:
\[ x-2=0 \quad\text{or}\quad x-3=0 \]
Therefore:
\[ x=2 \quad\text{or}\quad x=3 \]
Check
For \(x=2\):
\[ 2^2-5\cdot2+6=4-10+6=0 \]
For \(x=3\):
\[ 3^2-5\cdot3+6=9-15+6=0 \]
Both values work.
Key idea: For a quadratic \(x^2+bx+c=0\), try to write it as \((x+p)(x+q)=0\), where \(p+q=b\) and \(pq=c\). Then set each factor equal to zero.
Обсуждение
2 сообщенияHow does the sum-product trick work?
The sum–product trick works because of how two binomials multiply.
Start with:
\[ (x+p)(x+q) \]
Expand it:
\[ x^2+qx+px+pq=x^2+(p+q)x+pq \]
Now compare this with a quadratic written as:
\[ x^2+bx+c \]
Matching the coefficients gives:
\[ p+q=b \qquad\text{and}\qquad pq=c \]
So to factor \(x^2+bx+c\), you need two numbers whose sum is \(b\) and whose product is \(c\).
Example: \(x^2-5x+6=0\)
Here:
\[ b=-5,\qquad c=6 \]
We need two numbers that:
- add to \(-5\),
- multiply to \(6\).
Those numbers are \(-2\) and \(-3\), because:
\[ (-2)+(-3)=-5 \]
and
\[ (-2)(-3)=6 \]
So:
\[ x^2-5x+6=(x-2)(x-3) \]
Thus:
\[ x=2 \quad\text{or}\quad x=3 \]
Connection with the roots
If the roots are \(r\) and \(s\), then:
\[ x^2+bx+c=(x-r)(x-s) \]
Expanding:
\[ (x-r)(x-s)=x^2-(r+s)x+rs \]
So:
\[ r+s=-b \qquad\text{and}\qquad rs=c \]
That is Vieta’s formulas for a quadratic with leading coefficient \(1\): the roots have sum \(-b\) and product \(c\).
So the trick is just expanded-form matching: find the two numbers that fit both the sum and the product. If no convenient integer pair exists, the quadratic formula will always work.
