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The quadratic \(x^2 + \frac{3}{2}x -1\) has the following unexpected property: the roots, which are \(\frac{1}{2}\) and -2, are one less than the final two coefficients. Now find a quadratic with leading term \(x^2\) such that the final two coefficients are both non-zero, and the roots are one less than these coefficients.

 Feb 17, 2021
 #1
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Let a, b be the roots.

 

The the quadratic is x^2 - (a + b) x + ab.

 

Set a = -(a + b) - 1

b = ab - 1

 

Substituting, we have

b =b(-b - 2)/2 - 2

b = -b^2/2 - b - 2

b^2/2 + 2b + 2 = 0

b^2 + 4b + 4 = 0

(b + 2)^2 = 0

b = -2
 

Now, a = 0.

 

The required quadratic is x^2 - (0 + (-2)) x + 0(-2) = x^2 + 2x.

 Feb 17, 2021
 #2
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(x-0)(x-1)=0          roots             = 0  and +1

x^2 -x  + 0              coefficients = -1   and  0 

 Feb 17, 2021
edited by ElectricPavlov  Feb 17, 2021

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