How do you find the roots, real and imaginary, of $y = (\frac{x}{5} - 1) (- 2 x + 5)$ $y=(x/5-1)(-2x+5)$ using the quadratic formula?

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Answer 1 · 2017-01-15T04:38:21

$x \in {5, \frac{5}{2}}$ $x in {5, 5/2}$

No need to use the quadratic formula to find the roots since the equation is already factored out. Equate y to 0 to find the roots.

$y = 0 = (\frac{x}{5} - 1) (- 2 x + 5)$ $y = 0 = (x/5 - 1)(-2x + 5)$

$\Rightarrow \frac{x}{5} - 1 = 0$ $=> x/5 -1 = 0$ , and/or $- 2 x + 5 = 0$ $-2x + 5 = 0$

$\Rightarrow \frac{x}{5} = 1 \Rightarrow x = 5$ $=> x/5 = 1 => x = 5$

$\Rightarrow - 2 x = - 5 \Rightarrow x = \frac{5}{2}$ $=> -2x = -5 => x = 5/2$

If you really need to use the quadratic formula

$y = 0 = (\frac{x}{5} - 1) (- 2 x + 5)$ $y = 0 = (x/5 - 1)(-2x + 5)$

$\Rightarrow 0 = - \frac{2}{5} x^{2} + 3 x - 5$ $=> 0 = -2/5x^2 + 3x - 5$

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Given the quadratic equation

$a x^{2} + b x + c = 0$ $ax^2 + bx + c = 0$ , its roots can be determined by the formula

$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $x = (-b +- sqrt(b^2 - 4ac))/(2a)$

$\Rightarrow x = \frac{- 3 \pm \sqrt{3^{2} - 4 \cdot (- \frac{2}{5}) (- 5)}}{2 (- \frac{2}{5})}$ $=> x = (-3 +- sqrt(3^2 - 4*(-2/5)(-5)))/(2(-2/5)$

Proceeding with the computation should give us the same result as with the one above.

How do you find the roots, real and imaginary, of y=(x5−1)(−2x+5)y=(x/5-1)(-2x+5) using the quadratic formula?