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Hi, can you please help me solve question 2 please

Hi, can you please help me solve question 2 please-example-1
User TDrudge
by
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1 Answer

2 votes

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.

User Tlentali
by
7.6k points

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