Question

The equality $b^2+5>9b+12$ is satisfied if

• A. $b>9$
• B. $b<1$
• C. $b>0$
• D. $b<0$

Answer 1

Answer: (A), (D)

The correct options are
A $b>9$
D $b<0$

$f=b^2+5-9b-12=b^2-9b-7\Rightarrow b^2-2\times \left(\frac{9}{2}\right)\times b+{\left(\frac{9}{2}\right)}^2-{\left(\frac{9}{2}\right)}^2-7\Rightarrow b^2-2\times \left(\frac{9}{2}\right)\times b+{\left(\frac{9}{2}\right)}^2-\frac{81}{4}-7\Rightarrow {(b-\frac{9}{2})}^2-\frac{109}{4}$

For $f>0$

${(b-\frac{9}{2})}^2-\frac{109}{4}>0{(b-\frac{9}{2})}^2>\frac{109}{4}\Rightarrow -\sqrt{\frac{109}{4}}>(b-\frac{9}{2})and(b-\frac{9}{2})>\sqrt{\frac{109}{4}}\Rightarrow -5.22>(b-4.5)and(b-4.5)>+5.22\Rightarrow -0.72>bandb>+9.72sob<0andb>9.72$