Question

Difficulty: Very hardQuadratic Equations and the Quadratic Formula

In the quadratic equation ax2+bx+c=0a x^2 + b x + c = 0, the coefficients aa, bb, and cc are real numbers, and a=23a = \frac{2}{3}. If one of the roots of this equation is x=12i2x = \frac{1}{2} - i\sqrt{2}, where i=1i = \sqrt{-1}, what is the value of the product bcbc?

  1. 1-1Answer
  2. B
    79-\frac{7}{9}
  3. C
    12-\frac{1}{2}
  4. D
    00
  5. E
    79\frac{7}{9}

Answer

1-1
For any quadratic equation with real coefficients, if a complex number is a root, its conjugate must also be a root. Thus, the two roots are x1=12i2x_1 = \frac{1}{2} - i\sqrt{2} and x2=12+i2x_2 = \frac{1}{2} + i\sqrt{2}. Using Vieta's formulas, the sum of the roots is x1+x2=1=bax_1 + x_2 = 1 = -\frac{b}{a}, which gives b=23b = -\frac{2}{3}. The product of the roots is x1x2=142i2=94=cax_1 x_2 = \frac{1}{4} - 2i^2 = \frac{9}{4} = \frac{c}{a}, which gives c=32c = \frac{3}{2}. The product bcbc is therefore (23)(32)=1(-\frac{2}{3})(\frac{3}{2}) = -1.

Step-by-Step Solution

1
Find the second root of the quadratic equation.
The second root is x2=12+i2x_2 = \frac{1}{2} + i\sqrt{2}.
Since the quadratic equation has real coefficients, complex roots must occur in conjugate pairs.
2
Calculate the sum of the roots to find the coefficient bb.
b=23b = -\frac{2}{3}
The sum of the roots is x1+x2=(12i2)+(12+i2)=1x_1 + x_2 = (\frac{1}{2} - i\sqrt{2}) + (\frac{1}{2} + i\sqrt{2}) = 1. By Vieta's formulas, x1+x2=bax_1 + x_2 = -\frac{b}{a}. Substituting a=23a = \frac{2}{3} gives 1=b2/3    b=231 = -\frac{b}{2/3} \implies b = -\frac{2}{3}.
3
Calculate the product of the roots to find the coefficient cc.
c=32c = \frac{3}{2}
The product of the roots is x1x2=(12i2)(12+i2)=(12)2(i2)2=142i2x_1 x_2 = (\frac{1}{2} - i\sqrt{2})(\frac{1}{2} + i\sqrt{2}) = (\frac{1}{2})^2 - (i\sqrt{2})^2 = \frac{1}{4} - 2i^2. Since i2=1i^2 = -1, the product is \frac{1}{4} - 2(-1) = \frac{9}{4}.ByVietasformulas,. By Vieta's formulas, x_1 x_2 = \frac{c}{a}.Substituting. Substituting a = \frac{2}{3}gives94=c2/3    c=32 gives \frac{9}{4} = \frac{c}{2/3} \implies c = \frac{3}{2}.
4
Calculate the product bcbc.
bc=1bc = -1
Multiplying the values of bb and cc gives bc=(23)(32)=1bc = (-\frac{2}{3})(\frac{3}{2}) = -1.

Key Concept

Relating the roots of a quadratic equation with real coefficients to its coefficients via Vieta's formulas and the Complex Conjugate Root Theorem.
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