Question

If one zero of the quadratic polynomial \((k-1)x^2 + kx + 1\) is -4 then the value of k is

• A. \(-\frac{5}{4}\)
• B. \(\frac{5}{4}\)
• C. \(-\frac{4}{3}\)
• D. \(\frac{4}{3}\)

Hint:

When substituting a negative number into an expression with exponents, always use parentheses to avoid calculation errors. For example, write \((-4)^2\) which is 16, not \(-4^2\) which is -16.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
If a certain value is a "zero" of a polynomial, it means that when you substitute this value for the variable (x), the entire polynomial evaluates to zero.

Step 2: Key Formula or Approach:
Let the polynomial be \(P(x) = (k-1)x^2 + kx + 1\).
Since -4 is a zero, we have \(P(-4) = 0\). We will substitute \(x = -4\) into the polynomial and solve the resulting equation for k.

Step 3: Detailed Explanation:
Substitute \(x = -4\) into the polynomial:
\[ P(-4) = (k-1)(-4)^2 + k(-4) + 1 = 0 \] Now, simplify the equation:
\[ (k-1)(16) - 4k + 1 = 0 \] Distribute the 16:
\[ 16k - 16 - 4k + 1 = 0 \] Combine like terms (terms with k and constant terms):
\[ (16k - 4k) + (-16 + 1) = 0 \] \[ 12k - 15 = 0 \] Now, solve for k:
\[ 12k = 15 \] \[ k = \frac{15}{12} \] Simplify the fraction by dividing the numerator and denominator by their greatest common divisor, which is 3.
\[ k = \frac{15 \div 3}{12 \div 3} = \frac{5}{4} \]

Step 4: Final Answer:
The value of k is \(\frac{5}{4}\). This corresponds to option (B).