Question

Find the quadratic equation whose roots

are
2/3
and -1/2

(1) 6.x2 + x + 2 = 0
-2) 6.1² – x-2=0
(3) 16x? -x+2=0​

Answer 1

Answer:

$\begin{gathered}\begin{gathered}\begin{gathered}\sf Trigonometry\: Table \\ \begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\boxed{\begin{array}{ |c |c|c|c|c|c|} \bf\angle A & \bf{0}^{ \circ} & \bf{30}^{ \circ} & \bf{45}^{ \circ} & \bf{60}^{ \circ} & \bf{90}^{ \circ} \\ \\ \rm sin A & 0 & \dfrac{1}{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3}}{2} &1 \\ \\ \rm cos \: A & 1 & \dfrac{ \sqrt{3} }{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{1}{2} &0 \\ \\ \rm tan A & 0 & \dfrac{1}{ \sqrt{3} }&1 & \sqrt{3} & \rm \infty \\ \\ \rm cosec A & \rm \infty & 2& \sqrt{2} & \dfrac{2}{ \sqrt{3} } &1 \\ \\ \rm sec A & 1 & \dfrac{2}{ \sqrt{3} }& \sqrt{2} & 2 & \rm \infty \\ \\ \rm cot A & \rm \infty & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } & 0\end{array}}}\end{gathered}\end{gathered}\end{gathered} \end{gathered}\end{gathered}\end{gathered}\end{gathered}$

Answer 2

Step-by-step explanation:

$sum \: of \: zeroes = \frac{2}{3} + (- \frac{1}{2} ) = \frac{2}{3} - \frac{1}{2} = \frac{4 - 3}{2} = \frac{1}{2}$

$product \: of \: zeroes = \frac{2}{3} \times ( - \frac{1}{2} ) = - \frac{2}{6}$

$Quadratic \: equation = x² - (alpha + beta)x + alpha \: x \: beta$

$= x² - (1/2)x + (-2/6)$

$= 6x² - 3x -2$