Question

Topic :- Quadratic

Answer 1

The least integral value of 'm' for which the expression
mx^2 -4x +3m +1 is positive for every
x belongs to R is

Answer 2

just tell me what to do and why

Answer 3

So I understand the question correctly: You want to find \(m\) such that \(mx^2-4x+(3m+1)>0\), correct?

Answer 4

yup

Answer 5

So is it that D<0

Answer 6

i am just making a educated guess

Answer 7

Are you comfortable with a calculus method? It involves finding the vertex, which would be the minimum of the parabola if \(m>0\).

Answer 8

Nope, i know calculus , but i have to do it without calculus , though i cannot resist using it

Answer 9

That's alright, you can still find the vertex sans calculus. Complete the square and write in vertex form:
\[\begin{align*}mx^2-4x+(3m+1)&=m\left(x^2-\frac{4}{m}x+\frac{3m+1}{m}\right)\\
&=m\left(x^2-\frac{4}{m}x+\frac{4}{m^2}-\frac{4}{m^2}+\frac{3m+1}{m}\right)\\
&=m\left(\left(x-\frac{2}{m}\right)^2+\frac{3m^2+m-4}{m^2}\right)\\
&=m\left(x-\frac{2}{m}\right)^2+\frac{3m^2+m-4}{m}
\end{align*}\]
What's the vertex?

Answer 10

-b/2a

Answer 11

That's the x-coordinate, yes, but there's no need to compute all that stuff. Vertex form is \(a(x-h)^2+k\), where \((h,k)\) is the vertex.

This means the vertex of the parabola (assuming \(m>0\)) is a minimum at
\[\left(\frac{2}{m},~\frac{3m^2+m-4}{m}\right)\]

Answer 12

ans is 2

Answer 13

Set the y-coordinate equal to 0, then you have
\[\begin{align*}\frac{3m^2+m-4}{m}&=0\\
3m^2+m-4&=0\\
(3m+4)(m-1)&=0\\
m&=1,~-\frac{4}{3}
\end{align*}\]
We ignore the negative root. When \(m=1\), that makes the parabola touch the x-axis. We want the next integer, which would be \(m=2\). This guarantees that the parabola is positive for all \(x\). Makes sense?

Answer 14

Perfect!!!!!!!!!!!!!

Answer 15

thank you

Answer 16

You're welcome!

Answer 17

At 1 also it would be positive

Answer 18

and that is the least , so why not 1

Answer 19

Not quite. It'd be *mostly* positive, but zero at its vertex.
\[x^2-4x+4=(x-2)^2~~\Rightarrow~~\text{vertex at }(2,0)\]

Answer 20

but 1 is an integer and it is least than 2

Answer 21

Is there a concrete reason

Answer 22

Right, the fact that 1 is an integer and \(1<2\) is not up for debate here. It's the fact that when \(m=1\), the parabola is no longer positive everywhere. There's a difference between "being positive everywhere" and "not being negative everywhere." It comes down to the difference between \(\text{number}>0\) (positive) and \(\text{number}\ge0\) (non-negative).

Answer 23

Just a last question how do you decide that , it is positve everyhwhere or no

Answer 24

With parabolas, it's a question of where the vertex lies. The leading coefficient tells you whether the parabola opens up or down:

Clearly, the parabola will consist of only positive values if \(a>0\), which is why we assume \(m>0\).

The y-coordinate of the vertex must be greater than 0 for all values of \(x\) for the parabola to be positive everywhere.