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

Answer 1

#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#

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.

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Answer 2

To find the roots of the given quadratic equation ( y = \left(\frac{x}{5} - 1\right)(-2x + 5) ) using the quadratic formula:

  1. Rewrite the equation in the standard form ( ax^2 + bx + c = 0 ): ( y = -\frac{2}{5}x^2 + \frac{12}{5}x - 5 )

  2. Identify the coefficients: ( a = -\frac{2}{5} ), ( b = \frac{12}{5} ), ( c = -5 ).

  3. Apply the quadratic formula: ( x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}} ).

  4. Substitute the coefficients into the formula: ( x = \frac{{-\frac{12}{5} \pm \sqrt{{\left(\frac{12}{5}\right)^2 - 4 \cdot \left(-\frac{2}{5}\right) \cdot (-5)}}}}{{2 \cdot \left(-\frac{2}{5}\right)}} )

  5. Simplify: ( x = \frac{{-\frac{12}{5} \pm \sqrt{{\frac{144}{25} - \frac{40}{5}}}}}{{-\frac{4}{5}}} ) ( x = \frac{{-\frac{12}{5} \pm \sqrt{{\frac{144}{25} - \frac{200}{25}}}}}{{-\frac{4}{5}}} ) ( x = \frac{{-\frac{12}{5} \pm \sqrt{{-\frac{56}{25}}}}}{{-\frac{4}{5}}} )

  6. Simplify further: ( x = \frac{{-\frac{12}{5} \pm \frac{{\sqrt{56}}}{5}i}}{{-\frac{4}{5}}} ) ( x = \frac{{-12 \pm \sqrt{56}i}}{{-4}} ) ( x = \frac{3}{2} \pm \frac{{\sqrt{14}i}}{2} )

Therefore, the roots of the equation ( y = \left(\frac{x}{5} - 1\right)(-2x + 5) ) are ( x = \frac{3}{2} + \frac{{\sqrt{14}i}}{2} ) and ( x = \frac{3}{2} - \frac{{\sqrt{14}i}}{2} ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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