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.

eramsby1010 May 14, 2024

#1**0 **

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 −32c−32+c+1=c which simplifies to 31c=35 and so c=5.

Therefore, the quadratic equation is x2−32x+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.

cooIcooIcooI17 May 26, 2024