Let x2−(m−3)x+m=0,(m∈R) a quadratic equation. Find the value of m for which at least one root is greater than 2

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Cody Hart · Accepted Answer

Step 1For the existence of the roots, we need m≤1 or m≥9The roots are(m−3)±(m−3)2−4m2We want the larger root to be more than 2,m−3+(m−3)2−4m2&amp;gt;2vm2−10m+9&amp;gt;7−mClearly, any m≥9 would satisties the inequality as the RHS is negative and LHS is positive.If m≤1,m2−10m+9&amp;gt;m2−14m+49m&amp;gt;10and we find that the intersection is empty.Summary: m≥9

Mercedes Chang · Accepted Answer

Step 1We have, two cases.Case 1: One root is smaller than 2 and the other is greater than 2.⇒ 2  lies between the roots.This gives us a condition that: f(2)&amp;lt;0Case 2: Both roots are greater than 2This gives us the condition that: f(2)&amp;gt;0Also, D≥0 and −b2a&amp;gt;2From graph, Vertex: (−b2a,−D4a)After solving:Case 1 gives m&amp;gt;10, and Case 2 gives m∈[9, 10)Taking union of both cases, we get:m∈[9, ∞)

Let \(\displaystyle{x}^{{2}}-{\left({m}-{3}\right)}{x}+{m}={0},{\left({m}\in{R}\right)}\) a quadratic equation. Find the

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