Question

Q.17. Find the value of a for which the curve $y=x^2+ax+25$ touches the x-axis.

Answer 1

Dear Student,

$$\begin{gathered} ifthecurvey=x^2+ax+25touchesthexaxis \\ thantheequation{x}^2+ax+25=0hasequalandrealroots \\ ThereforethediscriminantD=0 \\ \Rightarrow b^2-4ac=0 \\ \Rightarrow a^2-4\times 25=0 \\ \Rightarrow a^2-100=0 \\ \Rightarrow a^2=100 \\ \Rightarrow a=\pm \sqrt{100} \\ a=\pm 10 \end{gathered}$$
Regards,