Value of an algebric expression in a quadratic equation

Question

I came across such a problem:

Given the equation \begin{equation} x^2 + \sqrt{m} x + n = 0 .\tag{1} \end{equation} If it has two equal real roots, what is the value of $(m+1)(m-1) - 2(2n - 1)$?

This is what I have done: Since the quadratic equation has two equal roots, we have \begin{equation} m - 4n = 0 ,\tag{2} \end{equation} or \begin{equation} m = 4n .\tag{3} \end{equation} However, \begin{equation} (m+1)(m-1) - 2(2n - 1) = m^2 - 4n + 1 ,\tag{4} \end{equation} I tried many ways to manipulate eq. (4), but couldn't figure out its value. I don't know how eqs. (2) or (3) can help. Is there any way out?

The $+1$ in (4) should be $+2$. Then simply substitute for $m$ using (3), The resulting expression in $n$ is your answer. — Ethan Bolker, Nov 22 '21 at 15:54
The +1 is -1+2. I agree that you get an expression in only one of the parameters ($m$ or $n$). — Charlie Vanaret, Nov 22 '21 at 15:59
All you can deduce is $m>4n$ from the condition of real roots. That gives you $m^2-4n+1>16n^2-4n+1=(4n-(1/2))^2+(3/4).$ But I don't see how you solve the question without additional information. — Marius S.L., Nov 22 '21 at 16:01
@Bonnaduck 's answer is correct. There is no further simplification. — Ethan Bolker, Nov 22 '21 at 16:04
@Bonnaduck Thanks, I missed that. Then my inequality will be equality. — Marius S.L., Nov 22 '21 at 16:04

Bonnaduck · Accepted Answer · 2021-11-22T16:06:04.333

4

We have

\begin{align*} (m+1)(m-1)-2(2n-1)&=m^2-1-4n+2\\ &=m^2-4n+1\\ &=m^2-m+1 \end{align*}

There's no way to further "solve" this problem. There are infinitely many values for $m$ that satisfy the given conditions.

It should be noted that if we are working with a polynomial over the reals, then we must also add the condition that $m\geq 0$, since $\sqrt\cdot$ cannot input negative values.

edited Nov 22 '21 at 16:06

answered Nov 22 '21 at 16:00

Bonnaduck

4,058

Value of an algebric expression in a quadratic equation

1 Answers1