Let $x^2 −(m−3)x+m=0,(m\in R)$ be a quadratic equation. Find the value of $m$ for which at least one root is greater than $2$

Question

Let $x^2 −(m−3)x+m=0,(m\in R)$ be a quadratic equation. Find the value of $m$ for which at least one root is greater than $2$.

Discriminant$=m^2-10m+9\ge0\implies m\in(-\infty,1]\cup[9,\infty)$

And putting $m=0$, we get both roots less than $2$. Does that mean we can eliminate all the values of $m\le1?$ Why?

Putting $m=9,$ we get double root at $3$. Does that mean all values of $m\ge9$ can be accepted?

Answer 1

For the existence of the roots, we need $m\le 1$ or $m \ge 9$.

The roots are

$$\frac{(m-3) \pm \sqrt{(m-3)^2-4m}}{2}$$

We want the larger root to be more than $2$,

$$\frac{m-3+\sqrt{(m-3)^2-4m}}{2}>2$$

$$\sqrt{m^2-10m+9} > 7-m$$

Clearly, any $m \ge 9$ would satisties the inequality as the RHS is negative and LHS is positive.

If $m \le 1$, $$m^2-10m+9 > m^2-14m+49$$

$$m > 10$$

and we find that the intersection is empty.

Summary: $m \ge 9$.

Answer 2

We have, two cases.

Case $1$: One root is smaller than $2$ and the other is greater than $2$.

$\implies 2$ lies between the roots.

This gives us a condition that: $f(2)<0$

Case $2$: Both roots are greater than $2$

This gives us the condition that: $f(2)>0$

Also, $D\ge0$ and $\frac{-b}{2a}>2$

[From graph, Vertex: $(\frac{-b}{2a}, \frac{-D}{4a})$ ]

After solving:

Case $1$ gives $m>10$, and Case $2$ gives $m \in [9,10)$

Taking union of both cases, we get:

$$\boxed{m\in [9, \infty) }$$

Answer 3

If $(f(x)=Ax^2+Bx+C=0$ has at least one root $>2$, them There are two cases: (1) one root $>2$, (2) both roots $>2$ For $f(x)=x^2-(m-3)x+m=0$

Case 1: $B^2\ge 4AC \implies m\le 1~or~ m\ge 9$ and $f(2)<0 \implies m>10$ the overlap of the two conditions gives $n\in(10,\infty)$.

Case2: $B^2>4AC:m\le 1 ~or~ m\ge 9$ and $x_0=-B/(2A) >2 \implies (m-3)/2>2 \implies m>7$ and $f(2)>0 \implies 10-m>0 \implies m<10.$ The overlap of the three intervals is $m \in [9,10)$.

So the union of the two cases gives $m\ge 9.$

Answer 4

The quadratic polynomial $ \ x^2 − (m−3)x + m \ \ $ represents an "upward-opening" parabola, for which the "vertex form" is $ \ \left(x \ − \ \frac{ m−3}{2} \right)^2 \ + \ \left( m \ - \ \frac{ [m−3]^2}{4} \right) \ \ . $ It may also be noted that $ \ f(1) \ = \ 4 \ \ , $ independent of the value of $ \ m \ \ . $ So there can be no zeroes of the polynomial greater than $ \ 2 \ $ at the very least for when the vertex is at $ \ x \le 1 \ \Rightarrow \ \frac{ m−3}{2} \ \le \ 1 \ \Rightarrow \ m \ \le \ 5 \ \ . $

The vertex is tangent to or "below" the $ \ x-$axis for $$ m \ - \ \frac{ [m−3]^2}{4} \ \le \ 0 \ \ \Rightarrow \ \ 0 \ \le \ -4m \ + \ m^2 \ -6m \ + \ 9 \ \ = \ \ (m - 1)·(m - 9) $$ $$ \Rightarrow \ \ m \ \le \ 1 \ \ , \ \ m \ \ge \ 9 $$ (as also found by other respondents). We may thus discard the first inequality. As for the second, $ \ m \ = \ 9 \ $ gives us the polynomial $ \ \left(x \ − \ \frac{ 9−3}{2} \right)^2 \ = \ (x \ − \ 3 )^2 \ \ , $ which has a "double zero" at $ \ x = 3 \ \ $ and is therefore an admissible result. All larger values of $ \ m \ $ correspond to parabolas with vertices at $ \ x \ \ge \ 3 \ $ and $ \ y \ < \ 0 \ \ , $ so they always have an $ \ x-$intercept "to the right" of $ \ x = 3 \ \ . $ Hence, all of the quadratic polynomials $ \ x^2 − (m−3)x + m \ \ $ have at least one real zero greater than $ \ 2 \ $ for $ \ \mathbf{m \ \ge \ 9} \ \ . $

[In fact, we may note that for $ \ m = 10 \ \ , $ the polynomial becomes $ \ x^2 - 7x + 10 \ = \ (x - 2)·(x - 5) \ \ , $ so both zeroes are greater than $ \ 2 \ $ for $ \ 9 \ \le \ m \ < \ 10 \ \ . \ $ ]