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Difficulty: EasyQuadratic Equations and the Quadratic Formula

One of the solutions to the quadratic equation 0.5x2−x+c=00.5x^2 - x + c = 0 is 1+i1 + i, where i=−1i = \sqrt{-1} and cc is a constant. What is the value of cc?

  1. A
    0
  2. B
    1 - 2i
  3. 1Answer
  4. D
    -1
  5. E
    1 + i

Answer

The constant cc is equal to 11.
Substituting the given solution x=1+ix = 1 + i into the equation 0.5x2−x+c=00.5x^2 - x + c = 0 gives 0.5(1+i)2−(1+i)+c=00.5(1+i)^2 - (1+i) + c = 0. Since (1+i)2=1+2i+i2=1+2i−1=2i(1+i)^2 = 1 + 2i + i^2 = 1 + 2i - 1 = 2i, the equation simplifies to 0.5(2i)−1−i+c=00.5(2i) - 1 - i + c = 0, which is i−1−i+c=0i - 1 - i + c = 0. Simplifying this yields −1+c=0-1 + c = 0, so the value of cc is 11.

Step-by-Step Solution

1
Substitute the solution x=1+ix = 1 + i into the quadratic equation 0.5x2−x+c=00.5x^2 - x + c = 0.
0.5(1+i)2−(1+i)+c=00.5(1+i)^2 - (1+i) + c = 0
Since 1+i1 + i is a solution to the equation, substituting it for xx must satisfy the equation.
2
Expand the squared term (1+i)2(1+i)^2 using the binomial square formula and the identity i2=−1i^2 = -1.
(1+i)2=1+2i+i2=1+2i−1=2i(1+i)^2 = 1 + 2i + i^2 = 1 + 2i - 1 = 2i
Expanding (a+b)2=a2+2ab+b2(a+b)^2 = a^2 + 2ab + b^2 with a=1a=1 and b=ib=i, and replacing i2i^2 with −1-1 simplifies the term.
3
Substitute 2i2i back into the equation and simplify.
0.5(2i)−(1+i)+c=0⇒i−1−i+c=0⇒−1+c=00.5(2i) - (1+i) + c = 0 \Rightarrow i - 1 - i + c = 0 \Rightarrow -1 + c = 0
Multiplying 0.50.5 by 2i2i yields ii, distributing the negative sign across (1+i)(1+i) yields −1−i-1 - i, and combining like terms leaves −1+c=0-1 + c = 0.
4
Solve the linear equation for cc.
c=1c = 1
Adding 11 to both sides isolates the variable cc.

Key Concept

Solving quadratic equations with complex roots by substitution and algebraic manipulation.
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Topics

Quadratic Equations and the Quadratic Formula

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