Question

If 9x4 + 12x3 + 28x2 + ax+bis a perfect square. find the values of

a and b.​

Answer 1

Answer:

I hipe is answer helpful

Answer 2

Answer:

If $9 {x}^{4} + 12 {x}^{3} + 28 {x}^{2} + ax + b$ is a perfect square then the value of a = 16 and b = 16

Step-by-step explanation:

1) Step 1:

Given polynomial is of degree 4 and it is a perfect square. It must be a square of degree 2 polynomial.

Let us assume $m {x}^{2} + nx + p$ be the square root of given polynomial $9 {x}^{4} + 12 {x}^{3} + 28 {x}^{2} + ax + b$

This implies

$9 {x}^{4} + 12 {x}^{3} + 28 {x}^{2} + ax + b = {(m {x}^{2} + nx + p)}^{2}$

2) Step 2:

Using the identity

( a + b + c )² = a² + b² + c² + 2ab + 2bc + 2ac

$= {(m {x}^{2} )}^{2} + {(nx)}^{2} + {p}^{2} + 2(m {x}^{2} )(nx) + 2(nx)(p) + 2(m {x}^{2})(p)$

$= {m}^{2} {x}^{4} + {n}^{2} {x}^{2} + {p}^{2} + 2mn {x}^{3} + 2npx + 2mp {x}^{2}$

$= {m}^{2} {x}^{4} + 2mn {x}^{3} + {n}^{2} {x}^{2} + 2mp {x}^{2} + 2npx + {p}^{2}$

3) Step 3:

• On comparing coefficient of x⁴,

${m}^{2} = 9 \: therefore \: m = \pm3$

Taking the positive value m = 3

• Comparing the coefficient of x³,

$2mn \: = 12 \: therefore \: n \: = \frac{6}{m} = \frac{6}{3} = 2$

n = 2

• Comparing coefficient of x²,

$28 = {n}^{2} + 2mp$

$28 = 4 + 2 \times 3 \times p$

$28 = 4 + 6p$

$24 = 6p$

$p = 4$

So all the values are m = 3 , n = 2 , p = 4

Now using them we will find the values of a and b

4) Step 4:

• Comparing the coefficient of x ,

$a \: = 2np \: = 2 \times 2 \times 4 = 16$

• Comparing constant terms,

$b \: = {p}^{2} = {4}^{2} = 16$

So the value of a and b comes out to be 16

a = 16 and b = 16

So if $9 {x}^{4} + 12 {x}^{3} + 28 {x}^{2} + ax + b$ is a perfect square then values of a and b are 16 both.

Note : Here we considered the positive value of m, we can also consider it's negative value in that case n and p will also be negative but the final values of a and b will remain same.