Question

If the roots of the equation x² – 8x + a² – 6a = 0 are real, then 'a' lies in the interval ____________.

Answer 1

$$\begin{gathered} \mathrm{For}{x}^2-8x+a^2-6a=0 \\ \mathrm{roots}\mathrm{are}\mathrm{real} \\ \mathrm{Here}a=1,b=-8,c=a^2-6a \\ \mathrm{i}.\mathrm{e}.D\ge 0\left(\mathrm{Discriminant}\right) \\ \mathrm{i}.\mathrm{e}.b^2-4ac\ge 0 \\ \mathrm{i}.\mathrm{e}.64-4\left(1\right)\left(a^2-6a\right)\ge 0 \\ \mathrm{i}.\mathrm{e}.64-4a^2+24a\ge 0 \\ \mathrm{i}.\mathrm{e}.-4a^2+24a+64\ge 0 \\ \mathrm{Now},\mathrm{multiplying}\mathrm{by}-\frac{1}{4} \\ \mathrm{i}.\mathrm{e}.\frac{4a^2}{4}-\frac{24a}{4}-\frac{64}{4}\le 0 \\ \mathrm{i}.\mathrm{e}.a^2-6a-16\le 0 \\ \mathrm{i}.\mathrm{e}.a^2-8a+2a-16\le 0 \\ \mathrm{i}.\mathrm{e}.a\left(a-8\right)+2\left(a-8\right)\le 0 \\ \mathrm{i}.\mathrm{e}.\left(a+2\right)\left(a-8\right)\le 0 \end{gathered}$$

for a < –2, (a + 2) (a – 8) ≥ 0 and for a > 8, (a + 2) (a – 8) ≥ 0