Question

If x= -3 and x = 2/3 are solutions of quadratic equation mx²+7x²+n=0 . Find the values of m and n ​

Answer 1

Step-by-step explanation:

mx² + 7x + n = 0

The zeroes are -3 and 2/3.

We know that,

Sum of zeroes = -b/a

Product of zeroes = c/a

$\alpha = - 3 \\ \beta = \frac{2}{3} \\ \\ \alpha + \beta = \frac{ - b}{a} \\ ( - 3) + ( \frac{2}{3} ) = \frac{ - 7}{m } \\ \frac{2 - 9}{3} = \frac{7}{m} \\ \frac{ - 7}{3} = \frac{7}{m} \\ m = \frac{7 \times 3}{ - 7} = - 3$

$\alpha \beta = \frac{c}{a} \\ ( - 3)( \frac{2}{3} ) = \frac{n}{ m} \\ - 2 = \frac{n}{ - 3} \\ n = ( - 3)( - 2) = 6$

Answer 2

Answer:

So m=3 and n=-6

Step-by-step explanation:

mx²+7x+n=0

Here a=m,b=7 and c=n

Roots are -3 and 2/3

Sum of roots=-b/a

-3+2/3=-7/m

-9m+2m=-21

-7m=-21, m=3

Product of roots= n/m

So (-3)*(2/3)=n/m=n/3

-2=n/3

n=-6

So m=3 and n=-6