Question

Find the value of L and m for which the
following expression will become perfect
square x^4+ 4x^3+16x^2 +lx+m​

Answer 1

l = 4 , m = 1

Step-by-step explanation:

( x + 1 )⁴ = x⁴ + 4x³ + 6x² + 4x + 1 [equation no. 1]

If x⁴ + 4x³ + 6x² + lx + m is a perfect square [From equation no. 1]

then it should be l = 4 , m = 1

Answer 2

$\large\underline{\sf{Solution-}}$

$\sf \: Let \: {x}^{4} + {4x}^{3} + {16x}^{2} + lx + m = {(a {x}^{2} + bx + c)}^{2}$

$\sf \: = \: {a}^{2} {x}^{4} + {b}^{2} {x}^{2} + {c}^{2} + 2ab {x}^{3} + 2bcx + 2ac {x}^{2}$

$\sf \: = \: {a}^{2} {x}^{4} + 2ab {x}^{3}+ {b}^{2} {x}^{2} + 2ac {x}^{2}+ 2bcx + {c}^{2}$

$\sf \: = \: {a}^{2} {x}^{4}+2ab {x}^{3}+ ({b}^{2}+ 2ac) {x}^{2}+ 2bcx + {c}^{2}$

So,

On comparing we get

$\red{\rm :\longmapsto\: {a}^{2} = 1 \implies \: a = 1 }- - - (1)$

$\blue{\rm :\longmapsto\:2ab = 4 \implies \: 2 \times 1 \times b = 4 \implies \: b = 2}$

$\green{\rm :\longmapsto\: {b}^{2} + 2ac = 16}$

Put values of a and b, we get

$\green{\rm :\longmapsto\: {2}^{2} + 2(1)c = 16}$

$\green{\rm :\longmapsto\:4 + 2c = 16}$

$\green{\rm :\longmapsto\:2c = 16 - 4}$

$\green{\rm :\longmapsto\:2c = 12}$

$\green{\rm :\longmapsto\:c = 6}$

Now,

$\pink{\rm :\longmapsto\:l = 2bc \implies \: l = 2 \times 2 \times 6 = 24}$

and

$\purple{\rm :\longmapsto\:m = {c}^{2} \implies \: m = {6}^{2} = 36}$

Additional Information :-

More Identities to know:

• (a + b)² = a² + 2ab + b²

• (a - b)² = a² - 2ab + b²

• a² - b² = (a + b)(a - b)

• (a + b)² = (a - b)² + 4ab

• (a - b)² = (a + b)² - 4ab

• (a + b)² + (a - b)² = 2(a² + b²)

• (a + b)³ = a³ + b³ + 3ab(a + b)

• (a - b)³ = a³ - b³ - 3ab(a - b)