Question

In the following, determine whether the given quadratic equations have real roots and if so, find the roots :
$16x^2$ = 24x + 1

• A. $\text{not real}$
• B. $real,\frac{3\pm \sqrt{10}}{4}$
• C. $\frac{3\pm \sqrt{10}}{2}$
• D. $\frac{3\pm \sqrt{13}}{4}$

Answer 1

Answer: (B)

$real,\frac{3\pm \sqrt{10}}{4}$

$16x^2=24x+1$

$\Rightarrow$ $16x^2-24x-1=0$

$\Rightarrow$ $a=16,b=-24,c=-1$

$\Rightarrow$ $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$=\frac{-(-24)\pm \sqrt{(24{)}^2-4(16\left)\right(-1)}}{2\left(16\right)}$

$=\frac{24\pm \sqrt{576+64}}{32}$

$=\frac{24\pm \sqrt{640}}{32}$

$=\frac{24\pm 8\sqrt{10}}{32}$

$=\frac{3\pm \sqrt{10}}{4}$

$\therefore$ $x=\frac{3\pm \sqrt{10}}{4}$

$\therefore$ We can see roots we have got is real.