Question

For the function f(x) = x² + 3x – 10 solve the following.

Answer 1

The given function is

`f(x)=x^2+3x-10`

We need to find the solution of

`\begin{gathered} f(x)\leq0 \\ x^2+3x-10\leq0 \end{gathered}`

Since the sign of inequality is <=, then the solution will be closed intervales from the small number to the big number [a, b]

At first, factor the left side into 2 factors

`\begin{gathered} x^2=(x)(x) \\ -10=(-2)(5) \\ (x)(-2)+(x)(5)=-2x+5x=3x \end{gathered}`

Then the two factors are (x - 2) and (x + 5)

`\begin{gathered} x^2+3x-10=(x-2)(x+5) \\ (x-2)(x+5)\leq0 \end{gathered}`

Now, equate each factor by 0 to find the values of x

`\begin{gathered} x-2=0,x+5=0 \\ x-2+2=0+2,x+5-5=0-5 \\ x=2,x=-5 \end{gathered}`

Since the smallest number -s -5 and the greatest number is 2, then

`-5\leq x\leq2`

The solution should be a closed interval from -5 to 2

The solution is [-5, 2]

The answer is

A. The solution is [-5, 2]