Question

In which interval are there solutions to the quadratic inequality x^2+7>8x?A. (1,7)B. [1, infinity)C. (-infinity,1]D. [1,7]

Answer 1

We are given the following quadratic inequality.

`x^2+7\ge8x`

Let us convert it into quadratic form and then factorize the equation

`\begin{gathered} x^2+7\ge8x \\ x^2-8x+7\ge0 \end{gathered}`

So now we need two numbers such that their sum is equal to -8 and their product is equal to 7.

How about -7 and -1 ?

Sum = -7 - 1 = -8

Product = -7*-1 = 7

`\begin{gathered} x^2-8x+7\ge0 \\ (x-7)(x-1)\ge0 \end{gathered}`

Now there are few possible cases

`\begin{gathered} (x-7)\ge0\quad and\quad (x-1)\ge0 \\ x\ge7\quad and\quad x\ge1 \end{gathered}`

Since x => 1 meets the requirement of x => 7 so we take x => 1 form here.

The other possible case is

`\begin{gathered} (x-7)\le0\quad and\quad (x-1)\le0 \\ x\le7\quad and\quad x\le1 \end{gathered}`

Since x <= 7 meets the requirement of x <= 1 so we take x <= 7 form here.

So the solution becomes

`\begin{gathered} x\ge1\quad and\quad x\le7 \\ 1\le x\le7 \end{gathered}`

Finally, the solution in interval notation is written as

`\lbrack1,7\rbrack`

We use brackets for closed intervals (less, greater or equal than cases)

We use parenthesis for open intervals (less, greater than cases)