In copying a second degree equation, a student mistakenly wrote the constant term as 6 instead of -6. His answers were -2 and -3. What are the answers to the original equation?
The correct quadratic equation is ax2 + bx - 6 = 0 The erroneous quadratic equation is ax2 + bx + 6 = 0
From ax2 + bx + 6 = 0, x1 = -2 and x2 = -3 Product of roots: $x_1 \, x_2 = \dfrac{c}{a}$
$-2(-3) = \dfrac{6}{a}$
$a = 1$
Sum of roots: $x_1 + x_2 = -\dfrac{b}{a}$
$-2 - 3 = -\dfrac{b}{1}$
$b = 5$
Thus, the correct equation is x2 + 5x - 6 = 0 $(x - 1)(x + 6) = 0$ $x = 1 ~ \text{and} ~ 6$
Thank you
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The correct quadratic equation is ax2 + bx - 6 = 0
The erroneous quadratic equation is ax2 + bx + 6 = 0
From ax2 + bx + 6 = 0, x1 = -2 and x2 = -3
Product of roots: $x_1 \, x_2 = \dfrac{c}{a}$
$-2(-3) = \dfrac{6}{a}$
$a = 1$
Sum of roots: $x_1 + x_2 = -\dfrac{b}{a}$
$-2 - 3 = -\dfrac{b}{1}$
$b = 5$
Thus, the correct equation is x2 + 5x - 6 = 0
$(x - 1)(x + 6) = 0$
$x = 1 ~ \text{and} ~ 6$
Thank you
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