Question

A student multiplies (4+5i) (3-2i) incorrectly and obtains 12-10i what is the students mistakes

Answer 1

Answer:

1. In the multiplication of the imaginary parts, the student forgot to square of i. OR

2. The student has only multiplied the real parts and the imaginary parts.

Correct value $=22+7i$.

Step-by-step explanation:

The given expression is

$(4+5i)(3-2i)$

A student multiplies (4+5i) (3-2i) incorrectly and obtains 12-10i.

Student's mistake can be either 1 or second:

1. In the multiplication of the imaginary parts, the student forgot to square of i.

2. The student has only multiplied the real parts and the imaginary parts.

$(4+5i)(3-2i)=4\times 3+5\times (-2)i=12-10i$

Which is not correct. The correct steps are shown below.

Using distributive property, we get

$4(3-2i)+5i(3-2i)$

$4(3)+4(-2i)+5i(3)+5i(-2i)$

$12-8i+15i-10i^2$

$12+7i-10(-1)$       $[\because i^2=-1]$

$12+7i+10$

$22+7i$

Therefore, the correct value of $(4+5i)(3-2i)$ is $22+7i$.