Math, asked by mathsssss52, 8 days ago

z/2+z/3+z/4 =13 ,zis equal to​

Answers

Answered by MiracleBrain
2

Answer :

Given: z = 2 – 3i

Given: z = 2 – 3iTo Prove: z2 – 4z + 13 = 0

Given: z = 2 – 3iTo Prove: z2 – 4z + 13 = 0Taking LHS, z2 – 4z + 13

Given: z = 2 – 3iTo Prove: z2 – 4z + 13 = 0Taking LHS, z2 – 4z + 13Putting the value of z = 2 – 3i, we get

Given: z = 2 – 3iTo Prove: z2 – 4z + 13 = 0Taking LHS, z2 – 4z + 13Putting the value of z = 2 – 3i, we get(2 – 3i)2 – 4(2 – 3i) + 13

Given: z = 2 – 3iTo Prove: z2 – 4z + 13 = 0Taking LHS, z2 – 4z + 13Putting the value of z = 2 – 3i, we get(2 – 3i)2 – 4(2 – 3i) + 13= 4 + 9i2 – 12i – 8 + 12i + 13

Given: z = 2 – 3iTo Prove: z2 – 4z + 13 = 0Taking LHS, z2 – 4z + 13Putting the value of z = 2 – 3i, we get(2 – 3i)2 – 4(2 – 3i) + 13= 4 + 9i2 – 12i – 8 + 12i + 13= 9(-1) + 9

Given: z = 2 – 3iTo Prove: z2 – 4z + 13 = 0Taking LHS, z2 – 4z + 13Putting the value of z = 2 – 3i, we get(2 – 3i)2 – 4(2 – 3i) + 13= 4 + 9i2 – 12i – 8 + 12i + 13= 9(-1) + 9= - 9 + 9

Given: z = 2 – 3iTo Prove: z2 – 4z + 13 = 0Taking LHS, z2 – 4z + 13Putting the value of z = 2 – 3i, we get(2 – 3i)2 – 4(2 – 3i) + 13= 4 + 9i2 – 12i – 8 + 12i + 13= 9(-1) + 9= - 9 + 9= 0

Given: z = 2 – 3iTo Prove: z2 – 4z + 13 = 0Taking LHS, z2 – 4z + 13Putting the value of z = 2 – 3i, we get(2 – 3i)2 – 4(2 – 3i) + 13= 4 + 9i2 – 12i – 8 + 12i + 13= 9(-1) + 9= - 9 + 9= 0= RHS

Given: z = 2 – 3iTo Prove: z2 – 4z + 13 = 0Taking LHS, z2 – 4z + 13Putting the value of z = 2 – 3i, we get(2 – 3i)2 – 4(2 – 3i) + 13= 4 + 9i2 – 12i – 8 + 12i + 13= 9(-1) + 9= - 9 + 9= 0= RHS= 4z3 – 16z2 + 52z + 13z2 – 52z + 169

Given: z = 2 – 3iTo Prove: z2 – 4z + 13 = 0Taking LHS, z2 – 4z + 13Putting the value of z = 2 – 3i, we get(2 – 3i)2 – 4(2 – 3i) + 13= 4 + 9i2 – 12i – 8 + 12i + 13= 9(-1) + 9= - 9 + 9= 0= RHS= 4z3 – 16z2 + 52z + 13z2 – 52z + 169= 4z(z2 – 4z + 13) + 13 (z2 – 4z + 13)

Given: z = 2 – 3iTo Prove: z2 – 4z + 13 = 0Taking LHS, z2 – 4z + 13Putting the value of z = 2 – 3i, we get(2 – 3i)2 – 4(2 – 3i) + 13= 4 + 9i2 – 12i – 8 + 12i + 13= 9(-1) + 9= - 9 + 9= 0= RHS= 4z3 – 16z2 + 52z + 13z2 – 52z + 169= 4z(z2 – 4z + 13) + 13 (z2 – 4z + 13)= 4z(0) + 13(0) [from eq. (i)]

Given: z = 2 – 3iTo Prove: z2 – 4z + 13 = 0Taking LHS, z2 – 4z + 13Putting the value of z = 2 – 3i, we get(2 – 3i)2 – 4(2 – 3i) + 13= 4 + 9i2 – 12i – 8 + 12i + 13= 9(-1) + 9= - 9 + 9= 0= RHS= 4z3 – 16z2 + 52z + 13z2 – 52z + 169= 4z(z2 – 4z + 13) + 13 (z2 – 4z + 13)= 4z(0) + 13(0) [from eq. (i)]= 0

Given: z = 2 – 3iTo Prove: z2 – 4z + 13 = 0Taking LHS, z2 – 4z + 13Putting the value of z = 2 – 3i, we get(2 – 3i)2 – 4(2 – 3i) + 13= 4 + 9i2 – 12i – 8 + 12i + 13= 9(-1) + 9= - 9 + 9= 0= RHS= 4z3 – 16z2 + 52z + 13z2 – 52z + 169= 4z(z2 – 4z + 13) + 13 (z2 – 4z + 13)= 4z(0) + 13(0) [from eq. (i)]= 0= RHS

Given: z = 2 – 3iTo Prove: z2 – 4z + 13 = 0Taking LHS, z2 – 4z + 13Putting the value of z = 2 – 3i, we get(2 – 3i)2 – 4(2 – 3i) + 13= 4 + 9i2 – 12i – 8 + 12i + 13= 9(-1) + 9= - 9 + 9= 0= RHS= 4z3 – 16z2 + 52z + 13z2 – 52z + 169= 4z(z2 – 4z + 13) + 13 (z2 – 4z + 13)= 4z(0) + 13(0) [from eq. (i)]= 0= RHSHence Proved

Answered by sabar7
1

Answer:

Z = 12

Step-by-step explanation:

Let's solve your equation step-by-step.

z /2 + z /3 + z /4 =13

Step 1: Simplify both sides of the equation.

z /2 + z /3 + z /4 =13

1 /2 z+ 1 /3 z+ 1 /4 z=13

( 1 /2 z+ 1 /3 z+ 1 /4 z)=13(Combine Like Terms)

13 /12 z=13  

Step 2: Multiply both sides by 12/13.

( 12 /13 )*( 13 /12 z)=( 12 /13 )*(13)

z=12

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