Question

Find the value of 2a ^ 2 + 3b ^ 2 + 4c ^ 2 when a = 1 , b = - 1 and c = 0 .​

Answer 1

Answer:

4a

2

+9b

2

+16c

2

+12ab+24bc+16ca

We know,

(x+y+z)

2

=x

2

+y

2

+z

2

+2xy+2yz+2zx [Identity]

Here,

x=2a

y=3b

z=4c

Putting Values in Identity, We get :

⇒ (2a+3b+4c)

2

=(2a)

2

+(3b)

2

+(4c)

2

+2(2a)(3b)+2(3b)(4c)+2(4c)(2a)

=4a

2

+9b

2

+16c

2

+12ab+24bc+16ca

Answer 2

Answer:

5

Step-by-step explanation:

2a² + 3b² + 4c²

= 2 × (1)² + 3 ×(-1)² + 4 × (0)². [since a=1, b=-1, c=0]

= (2 × 1) + (3 × 1 ) + (4 × 0). [ (1)²=1, (-1)²= 1, 0²= 0 ]

= 2 + 3 + 0

= 5