Question

question
which quadratic equation has roots
$-7 + 2i$ and $-7 - 2i$
answer
$2x^{2}+28x - 106 = 0$
$2x^{2}-28x - 106 = 0$
$2x^{2}-28x + 106 = 0$
$2x^{2}+28x + 106 = 0$
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Explanation:

Step1: Recall the sum and product of roots

For a quadratic equation \(ax^2 + bx + c = 0\) with roots \(r_1\) and \(r_2\), the sum of roots \(r_1 + r_2=-\frac{b}{a}\) and the product of roots \(r_1\times r_2=\frac{c}{a}\). Here, \(r_1=-7 + 2i\) and \(r_2=-7 - 2i\).

Step2: Calculate the sum of the roots

\(r_1 + r_2=(-7 + 2i)+(-7 - 2i)=-14\)

Step3: Calculate the product of the roots

\(r_1\times r_2=(-7 + 2i)(-7 - 2i)=(-7)^2-(2i)^2 = 49-4i^2\). Since \(i^2=-1\), this becomes \(49 + 4=53\)

Step4: Form the quadratic equation

Let the quadratic equation be \(ax^2+bx + c = 0\). From the options, \(a = 2\).

• Sum of roots: \(-\frac{b}{a}=-14\), substituting \(a = 2\), we get \(-\frac{b}{2}=-14\), so \(b = 28\) (but since the sum is \(-14\) and the formula is \(-\frac{b}{a}\), if we consider the sign, actually, when we build the equation from roots, the equation is \(x^2-(r_1 + r_2)x+r_1r_2 = 0\). Let's first build the monic equation (\(a = 1\)): \(x^2-(-14)x + 53=0\) i.e., \(x^2 + 14x+53 = 0\). Now, to match the coefficient \(a = 2\) in the options, multiply the entire equation by 2: \(2x^2+28x + 106=0\)

Answer:

\(2x^2 + 28x + 106 = 0\) (the last option)