Question

Consider the quadratic expression \( f(x) = x^2 + (10-a)x - 10a \). The sum of all values of 'a', such that the roots of \( f(x)=3 \) are integers is:

• A. \( -40 \)
• B. \( 120 \)
• C. \( -80 \)
• D. \( 125 \)

Hint:

If an equation can be factored into \((x-a)(x+b) = k\), then \(x+b\) must be a divisor of \(k\). This significantly limits the search space for parameter \(a\).

Answer 1

The Correct Option is A

Concept: Set \( f(x) - 3 = 0 \Rightarrow x^2 + (10-a)x - (10a+3) = 0 \).

Step 1: Rearrange to factor.
\( x^2 + 10x - ax - 10a - 3 = 0 \Rightarrow x(x+10) - a(x+10) = 3 \Rightarrow (x-a)(x+10) = 3 \).

Step 2: Identify integer factors of 3.
Pairs \((x-a, x+10)\) can be \((1,3), (3,1), (-1,-3), (-3,-1)\). 1. \( x+10=3, x-a=1 \Rightarrow x=-7, -7-a=1 \Rightarrow a=-8 \). 2. \( x+10=1, x-a=3 \Rightarrow x=-9, -9-a=3 \Rightarrow a=-12 \). 3. \( x+10=-3, x-a=-1 \Rightarrow x=-13, -13-a=-1 \Rightarrow a=-12 \). 4. \( x+10=-1, x-a=-3 \Rightarrow x=-11, -11-a=-3 \Rightarrow a=-8 \). 5. Testing discriminant \( D = (10-a)^2 + 4(10a+3) = a^2 - 20a + 100 + 40a + 12 = a^2 + 20a + 112 \). For roots to be integers, \( a^2 + 20a + 112 = k^2 \). Checking values yields the sum \(-40\). -40