Question 513 Marks
Find the roots of the following equations, if they exist, by applying the quadratic formula:
$ 16 x^2=24 x+1 $
$ 16 x^2=24 x+1 $
Answer
View full question & answer→Given,
$ 16 x^2=24 x+1 $
$16 x^2-24 x-1=0$
On comparing it with $a x^2+b x+c=0$
$a=16, b=-24 \text { and } c=-1$
Discriminant $D$ is given by:
$ D=\left(b^2-4 a c\right) $
$ =(-24)^2-4 \times 16 \times(-1)$
$ =576+(64) $
$=640>0$
Hence, the roots of the equation are real,
Roots $\alpha$ and $\beta$ are given by:
$\alpha=\frac{-\text{b}+\sqrt{\text{D}}}{2\text{a}}$
$=\frac{-(-24)+\sqrt{640}}{2\times16}$
$=\frac{24+8\sqrt{10}}{32}$
$=\frac{8\big(3+\sqrt{10}\big)}{32}$
$=\frac{\big(3+\sqrt{10}\big)}{4}$
$\beta=\frac{-\text{b}-\sqrt{\text{D}}}{2\text{a}}$
$=\frac{-(-24)+\sqrt{640}}{2\times16}$
$=\frac{24-8\sqrt{10}}{32}$
$=\frac{8\big(3-\sqrt{10}\big)}{32}$
$=\frac{\big(3-\sqrt{10}\big)}{4}$
Thus, the roots of the equation are $\frac{\big(3+\sqrt{10}\big)}{4}$ and $\frac{\big(3-\sqrt{10}\big)}{4}.$
$ 16 x^2=24 x+1 $
$16 x^2-24 x-1=0$
On comparing it with $a x^2+b x+c=0$
$a=16, b=-24 \text { and } c=-1$
Discriminant $D$ is given by:
$ D=\left(b^2-4 a c\right) $
$ =(-24)^2-4 \times 16 \times(-1)$
$ =576+(64) $
$=640>0$
Hence, the roots of the equation are real,
Roots $\alpha$ and $\beta$ are given by:
$\alpha=\frac{-\text{b}+\sqrt{\text{D}}}{2\text{a}}$
$=\frac{-(-24)+\sqrt{640}}{2\times16}$
$=\frac{24+8\sqrt{10}}{32}$
$=\frac{8\big(3+\sqrt{10}\big)}{32}$
$=\frac{\big(3+\sqrt{10}\big)}{4}$
$\beta=\frac{-\text{b}-\sqrt{\text{D}}}{2\text{a}}$
$=\frac{-(-24)+\sqrt{640}}{2\times16}$
$=\frac{24-8\sqrt{10}}{32}$
$=\frac{8\big(3-\sqrt{10}\big)}{32}$
$=\frac{\big(3-\sqrt{10}\big)}{4}$
Thus, the roots of the equation are $\frac{\big(3+\sqrt{10}\big)}{4}$ and $\frac{\big(3-\sqrt{10}\big)}{4}.$