Question

9x^4+12x^3+28x^2+(9m+7n)x+7n+9m is a perfect square. Then find m,n value

Answer 1

Answer:

The integer single digit value of m = 8, n = -8

Step-by-step explanation:

Given expression

$9x^4+12x^3+28x^2+(9m+7n)x+(9m+7n)$

We know that

$(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca$

Therefore in order to make the expression a perfect square, rearranging it

$9x^4+12x^3+28x^2+(9m+7n)x+(9m+7n)$

$=(3x^2)^2+2\times 2\times (3x^2)+24x^2+4x^2+(9m+7n)x+(9m+7n)$

$=(3x^2)^2+2\times 2x\times (3x^2)+2\times (3x^2)\times 4+(2x)^2+(9m+7n)x+(9m+7n)$

This means

$a=3x^2\\b=2x\\c=4$

Therefore we can say that the value of

9m + 7n = 16           ................... (1)

And the value of

9m + 7n = 16            ................... (2)

Equations (1) and (2) are essentially the same

m = (16 - 7n)/9

   = (9 + 7 - 7n)/9

   = 1 + 7(1 - n)/9

For m and n to be integers 1 - n should be divisible by 9

Miniumum positive value of n = 10

Therefore value of m = -6

Minimum negative value of n = -8

Then value of m = 8