If `10a^(2)-27a+5=0`, find values of a

Answer

To solve the quadratic equation \(10a^2 - 27a + 5 = 0\), we will follow these steps:

### Step 1: Identify the coefficients
The given quadratic equation is in the standard form \(Ax^2 + Bx + C = 0\), where:
- \(A = 10\)
- \(B = -27\)
- \(C = 5\)

### Step 2: Calculate the product of \(A\) and \(C\)
We need to find the product of the coefficient of \(a^2\) (which is \(A\)) and the constant term (which is \(C\)):
\[ A \cdot C = 10 \cdot 5 = 50 \]

### Step 3: Find two numbers that multiply to \(A \cdot C\) and add to \(B\)
We need to find two numbers that multiply to \(50\) and add up to \(-27\). The numbers are \(-25\) and \(-2\) because:
\[ -25 \cdot -2 = 50 \quad \text{and} \quad -25 + -2 = -27 \]

### Step 4: Rewrite the middle term
Now, we can rewrite the equation by splitting the middle term using the two numbers we found:
\[ 10a^2 - 25a - 2a + 5 = 0 \]

### Step 5: Factor by grouping
Next, we will group the terms:
\[ (10a^2 - 25a) + (-2a + 5) = 0 \]
Now, factor out the common factors:
\[ 5a(2a - 5) - 1(2a - 5) = 0 \]
This can be rewritten as:
\[ (5a - 1)(2a - 5) = 0 \]

### Step 6: Set each factor to zero
Now, we set each factor equal to zero:
1. \(5a - 1 = 0\)
2. \(2a - 5 = 0\)

### Step 7: Solve for \(a\)
Solving the first equation:
\[ 5a - 1 = 0 \implies 5a = 1 \implies a = \frac{1}{5} \]

Solving the second equation:
\[ 2a - 5 = 0 \implies 2a = 5 \implies a = \frac{5}{2} \]

### Final Answer
The values of \(a\) are:
\[ a = \frac{1}{5} \quad \text{and} \quad a = \frac{5}{2} \]

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