Solve the following inequalities: (a) x²+x-6 (b) x²+8x+1

Question

Solve the following inequalities:

(a) x²+x-6 (b) x²+8x+1<0

Accepted Answer

[latex]\begin{array}{r}(a)\quad x^2+x-6>0 \ (x-2)(x+3)>0 \end{array}[/latex]

For the product (x — 2)(x + 3) to be positive requires either

(i) [latex]x-2>0 \text { and } x+3>0[/latex]

or

(ii) [latex]x-2<0 \text { and } x+3<0[/latex]

Case (i) [latex]\begin{aligned} & x-2>0 \text { and so } x>2 \ & x+3>0 \text { and so } x>-3 \end{aligned}[/latex]

Both of these inequalities are satisfied only when x > 2.

Case (ii) [latex]\begin{aligned} & x-2<0 \text { and so } x<2 \& x+3<0 \text { and so } x<-3 \end{aligned}[/latex]

Both of these inequalities are satisfied only when x < —3.
In summary, x² + x — 6 > 0 when either x > 2 or x < —3.

(b) The quadratic expression x² + 8x + 1 does not factorize and so the technique of completing the square is used.

[latex]x^2+8 x+1=(x+4)^2-15[/latex]

Hence

[latex]\begin{aligned} (x+4)^2-15 & <0 \(x+4)^2 & <15\end{aligned}[/latex]

Using the result after Example 1.36 we may write

[latex]\begin{aligned} -\sqrt{15} &

Question 1.37: Solve the following inequalities: (a) x²+x-6 (b) x²+8x+1&......