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

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Answer 1 · 2017-01-15T04:38:21

$x in {5, 5/2}$

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.

$y = 0 = (x/5 - 1)(-2x + 5)$

$=> x/5 -1 = 0$ , and/or $-2x + 5 = 0$

$=> x/5 = 1 => x = 5$

$=> -2x = -5 => x = 5/2$

If you really need to use the quadratic formula

$y = 0 = (x/5 - 1)(-2x + 5)$

$=> 0 = -2/5x^2 + 3x - 5$

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Given the quadratic equation

$ax^2 + bx + c = 0$ , its roots can be determined by the formula

$x = (-b +- sqrt(b^2 - 4ac))/(2a)$

$=> x = (-3 +- sqrt(3^2 - 4*(-2/5)(-5)))/(2(-2/5)$

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=(x/5-1)(-2x+5)y=(x/5-1)(-2x+5) using the quadratic formula?