# What are the Roots of the Original Quadratic Equation

3 posts / 0 new
Noufiya Rasheed
What are the Roots of the Original Quadratic Equation

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?

Jhun Vert

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
$(x - 1)(x + 6) = 0$
$x = 1 ~ \text{and} ~ 6$

Noufiya Rasheed

Thank you

• Mathematics inside the configured delimiters is rendered by MathJax. The default math delimiters are $$...$$ and $...$ for displayed mathematics, and $...$ and $...$ for in-line mathematics.