Question

For what values of m does the graph of y = 3x2 + 7x + m have two x-intercepts

Answer 1

we are given

$y=3x^2+7x+m$

To find x-intercept means we have to find zeros

and for finding zeros , we will use quadratic formula

and we have it has two x-intercepts

so, it's discriminant must be greater than 0

so, we will find discriminant

$D=\sqrt{b^2-4ac}$

now, we can compare with

$y=ax^2+bx+c$

$y=3x^2+7x+m$

and then we can find a , b and c

$a=3,b=7,c=m$

now, we can find D

$D=\sqrt{7^2-4*3*m}$

$D=\sqrt{49-12m}$

It has two x-intercepts

so,

$D=\sqrt{49-12m}>0$

now, we can solve for m

$49-12m>0$

$12m$

$m$............Answer

Answer 2

Answer:

$m < \frac{49}{12}$

Step-by-step explanation:

$y= 3x^2+7x+m$

a = 3

b=7

c = m

The graph is a parabola .

The Parabola with minimum point $x=\frac{-b}{2a}$  

                                                         $x=\frac{-7}{2\times 3}$  

                                                          $x=\frac{-7}{6}$                                                            

Now, To have 2 x intercepts, the value at the minimum point must be less than 0 i.e. $3x^2+7x+m$

So, Substitute the minimum value in the equation

$3(\frac{49}{36}) - \frac{49}{6}+ m < 0$

$\frac{49}{12} - \frac{49}{6}+ m < 0$

$\frac{49-98}{12}+ m < 0$

$\frac{-49}{12}+ m < 0$

$m < \frac{49}{12}$

Thus  the graph of $y= 3x^2+7x+m$ have two x-intercepts when $m < \frac{49}{12}$