Question 513 Marks
Find the roots of the following equations, if they exist, by applying the quadratic formula:
$16x^2 = 24x + 1$
$16x^2 = 24x + 1$
Answer
View full question & answer→Given,
$16x^2 = 24x + 1$
$16x^2 - 24x - 1 = 0$
On comparing it with $ax^2 + bx + c = 0$
$a = 16, b = -24$ and c = -1
Discriminant D is given by:
$D = (b^2 - 4ac)$
$= (-24)^2 - 4 \times 16 \times (-1)$
$= 576 + (64)$
$= 640 > 0$
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 $\frac{\big(3-\sqrt{10}\big)}{4}.$
$16x^2 = 24x + 1$
$16x^2 - 24x - 1 = 0$
On comparing it with $ax^2 + bx + c = 0$
$a = 16, b = -24$ and c = -1
Discriminant D is given by:
$D = (b^2 - 4ac)$
$= (-24)^2 - 4 \times 16 \times (-1)$
$= 576 + (64)$
$= 640 > 0$
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 $\frac{\big(3-\sqrt{10}\big)}{4}.$