Question
which of the following quadratic equations is not solvable by grouping? x^2-12x+35=0
2x^2-2x-10=0
x^2-2x+1=0
2x^2+14x+12=0
2x^2-2x-10=0
x^2-2x+1=0
2x^2+14x+12=0
Answers
GPT-4o mini
To determine which of the given quadratic equations is not solvable by grouping, we can analyze each equation.
**1. \(x^2 - 12x + 35 = 0\)**
To factor by grouping, we look for two numbers that multiply to \(35\) and add to \(-12\). Those numbers are \(-5\) and \(-7\).
Thus, \(x^2 - 12x + 35 = (x - 5)(x - 7) = 0\). This equation is solvable by grouping.
**2. \(2x^2 - 2x - 10 = 0\)**
First, we can factor out a common \(2\):
\[ 2(x^2 - x - 5) = 0 \]
Now, we need to find two numbers that multiply to \(-5\) and add to \(-1\). Since there are no integer pairs that satisfy this condition, this equation likely cannot be factored simply or solved by grouping directly.
**3. \(x^2 - 2x + 1 = 0\)**
This is a perfect square:
\[ (x - 1)^2 = 0 \]
This equation is also solvable by grouping.
**4. \(2x^2 + 14x + 12 = 0\)**
Again, we can factor out a common \(2\):
\[ 2(x^2 + 7x + 6) = 0 \]
Now we look for two numbers that multiply to \(6\) and add to \(7\), which are \(6\) and \(1\):
\[ x^2 + 7x + 6 = (x + 6)(x + 1) = 0 \]
So this equation is also solvable by grouping.
From this analysis, the quadratic equation which is **not solvable by grouping** is:
**\(2x^2 - 2x - 10 = 0\)**.
**1. \(x^2 - 12x + 35 = 0\)**
To factor by grouping, we look for two numbers that multiply to \(35\) and add to \(-12\). Those numbers are \(-5\) and \(-7\).
Thus, \(x^2 - 12x + 35 = (x - 5)(x - 7) = 0\). This equation is solvable by grouping.
**2. \(2x^2 - 2x - 10 = 0\)**
First, we can factor out a common \(2\):
\[ 2(x^2 - x - 5) = 0 \]
Now, we need to find two numbers that multiply to \(-5\) and add to \(-1\). Since there are no integer pairs that satisfy this condition, this equation likely cannot be factored simply or solved by grouping directly.
**3. \(x^2 - 2x + 1 = 0\)**
This is a perfect square:
\[ (x - 1)^2 = 0 \]
This equation is also solvable by grouping.
**4. \(2x^2 + 14x + 12 = 0\)**
Again, we can factor out a common \(2\):
\[ 2(x^2 + 7x + 6) = 0 \]
Now we look for two numbers that multiply to \(6\) and add to \(7\), which are \(6\) and \(1\):
\[ x^2 + 7x + 6 = (x + 6)(x + 1) = 0 \]
So this equation is also solvable by grouping.
From this analysis, the quadratic equation which is **not solvable by grouping** is:
**\(2x^2 - 2x - 10 = 0\)**.