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 more than these coefficients.

 May 14, 2024

Let the quadratic be of the form x2+bx+c=0. We are given that the roots (solutions) are one more than the final two coefficients, b and c. Let the roots be b+1 and c+1.


We can use Vieta's formulas to relate the coefficients of a quadratic equation to its roots. Vieta's formulas state that the sum of the roots is equal to the negative of the coefficient of the x term, and the product of the roots is equal to the constant term.


In this case, the sum of the roots is: (b+1)+(c+1)=−b which simplifies to b+c=−b−2 and so 2b=−b−2 which implies b=−32​.


The product of the roots is: (b+1)⋅(c+1)=c, which simplifies to bc+b+c+1=c. Since b=−32​, this becomes −32​c−32​+c+1=c which simplifies to 31​c=35​ and so c=5.


Therefore, the quadratic equation is x2−32​x+5=0. We can check that the roots of this equation are indeed 32​+1=35​ and 5+1=6. So, the quadratic we seek is x^2−2/3*x+5.

 May 26, 2024

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