Question

Hi, can you please help me solve question 2 please

Answer 1

1.

We need to find the roots of the next quadratic function:

`z^2+8z+65`

Use the quadratic formula, which is given by the form ax²+bx+c:

`\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}`

Replace the values using a=1, b=8 and c=65

`\frac{-8\pm\sqrt[]{(8)^2-4(1)(65)}}{2(1)}`
`\frac{-8\pm\sqrt[]{64-260}}{2}=\frac{-8\pm\sqrt[]{-196}}{2}=(-8\pm14i)/(2)`

Simplify the expression:

`(-8\pm14i)/(2)=(2(-4\pm7i))/(2)=-4\pm7i`

Therefore, the roots of the given expression are:

x = -4+7i

x= -4 - 7i

2.

We need to find the quadratic using the integer coefficients:

We have that the root are :

`(1)/(2)\pm(3)/(2)i`

Use the quadratic form ax²+bx+c

The roots are the solution for the quadratic formula, so:

`\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}=(1)/(2)\pm(3)/(2)i`

Then:

`\frac{1\pm\sqrt[]{9i^2}}{2}=\frac{1\pm\sqrt[]{-9}}{2}`

So we have that 2a =2. then a =2/2

a=1

Also, -b = 1

b = -1

b²-4ac = -9

(-1)²-4(1)c = -9

-4(1)c = -9 -1

-4c = -10

Solve for c

c = -10/-4

c = 5/2

Therefore, with these values, we can replace the quadratic form and we will find the result

a=1

b = -1

c = 5/2

`ax^2+bx+c\text{ =0}`
`x^2-x+(5)/(2)=0`

So, the last quadratic is the result.