Question

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Answer 1

Question: If α and β are the roots of the quadratic equation 2x² + 3x - 7 = 0, then find the values of (α² + β²)/αβ.

Solution:

Since we have been given the quadratic equation, we'll try to find the value of α + β and αβ with the help of the relation between zeroes and co-efficients using the below formulas,

$\sf p(x) = ax^2 + bx + c\\ \\\sf 'a' \ is \ the \ coefficient \ of \ x^2.\\ \\\sf 'b' \ is \ the \ coefficient \ of \ x.\\ \\\sf 'c' \ is \ the \ constant.\\ \\ \\If \ \alpha \ and \ \beta \ are \ the \ zeroes,\\ \\ \\\Longrightarrow \sf \alpha + \beta = \dfrac{-b}{a}\\ \\ \\\Longrightarrow \sf \alpha \beta = \dfrac{c}{a}$

ATQ,

$\displaystyle \sf p(x) = 2x^2 + 3x - 7\\ \\Here, \sf a = 2, b = 3, and c = -7 \\ \\ \\\Longrightarrow \sf \alpha + \beta = \dfrac{-b}{ \ \ a}\\ \\ \\\Longrightarrow \sf \alpha + \beta = \dfrac{-3}{ \ \ 2}\ \ \ \longmapsto \textcircled{\scriptsize {1}}$

$\Longrightarrow \sf \alpha \beta = \dfrac{c}{a}\\ \\ \\ \Longrightarrow \sf \alpha \beta = \dfrac{-7}{2}\ \ \ \longmapsto \textcircled{\scriptsize \sf {2}}$

$\sf TO \ FIND: \dfrac{\alpha^2+\beta^2}{\alpha\beta}\\ \\ \\\Longrightarrow \sf \dfrac{\alpha^2+\beta^2}{\alpha\beta}\\ \\ \\\boxed{\begin{minipage}{7 cm}\sf \ \ \ \ We \ know \ that (a+b)^2= \sf a^2+b^2+2ab \\ \\\sf { \ \ \ \ \ \ \ Similarly, (a+b)^2-\sf 2ab=a^2+b^2}\end{minipage}}\\ \\ \\ \\\Longrightarrow\dfrac{(\alpha+\beta)^2-2\alpha\beta}{\alpha\beta}$

On substituting 1 and 2 below we get.

$\Longrightarrow \sf \dfrac{\left(\dfrac{-3}{2}\right)^2-2\left(\dfrac{-7}{2}\right)}{\left(\dfrac{-7}{ \ \ 2}}\right)}}\\ \\ \\\Longrightarrow \sf \dfrac{\left(\dfrac{9}{4}\right)+7}{\left(\dfrac{-7}{ \ \ 2}\right)}\\ \\ \\ \\\Longrightarrow \sf \dfrac{\left(\dfrac{9+28}{4}\right)}{\left(\dfrac{-7}{ \ \ 2\right)}}\\ \\ \\ \\\Longrightarrow \sf \dfrac{\left(\dfrac{37}{4}\right)}{\left(\dfrac{-7}{ \ \ 2\right)}}\\ \\ \\ \\\Longrightarrow \sf \dfrac{37}{4} \times \dfrac{2}{-7}$

$\Longrightarrow \sf \ \dfrac{37 \times 2}{4 \times -7}\\ \\ \\\Longrightarrow \sf \ \dfrac{74}{-28}\\ \\ \\\Longrightarrow \sf \ -\dfrac{37}{14}$

$\boxed{\boxed{ \ \therefore \ \sf \frac{\left(\alpha + \beta\right)^2}{\alpha \beta} = -\frac{37}{14} \ }}$

Answer 2

$\bf Given:\ \tt \alpha\ and\ \beta\ \are\ the\ roots\ of\ the\ equation\ 2x^2\ +\ 3x\ -\ 7.\\\\\\\bf To\ find:\ \tt\ The\ value\ of\ \dfrac{\alpha^2\ +\ \beta^2}{\alpha\beta}.\\\\\\\bf Answer:$

$\tt Let's\ suppose\ ax^2\ +\ bx\ +\ c\ =\ 0\ is\ \quadratic\ equation.\\\\\\The\ sum\ of\ its\ roots\ would\ be\ given\ by:\ \dfrac{-b}{a}.\\\\\\The\ product\ of\ its\ roots\ would\ be\ given\ by:\ \dfrac{c}{a}.\\\\\\From\ the\ equation\ given\ to\ us,\\\\\\a\ =\ 2,\ b\ =\ 3\ and\ c\ =\ -7.$

$\tt It's\ given\ that\ \alpha\ and\ \beta\ are\ the\ roots.\\\\\\\implies\ \alpha\ +\ \beta\ =\ =\ \dfrac{-b}{a}\ =\ \dfrac{-3}{2}.\\\\\\\implies\ \alpha\ \times\ \beta\ =\ \dfrac{c}{a}\ =\ \dfrac{-7}{2}.$

$\tt Now,\ we\ have\ to\ find\ the\ value\ of\ \dfrac{\alpha^2\ +\ \beta^2}{\alpha\beta}.\\\\\\We\ have\ the\ value\ of\ \alpha\beta\ but\ not\ \alpha^2\ +\ \beta^2.$

$\tt \bigg[Through, the\ (a\ +\ b)^2\ identity,\ we\ can\ say\ that\ \\ \\\alpha^2\ +\ \beta^2\ =\ (\alpha\ +\ \beta)^2\ -\ 2\alpha\beta\bigg]\\\\\\Using\ the\ values\ we\ have,\\\\\\\implies\ \dfrac{(\alpha\ +\ \beta)^2\ -\ 2\alpha\beta}{\alpha\beta}\\\\\\\implies\ \dfrac{\bigg(\dfrac{-3}{2}\bigg)^2\ -\ 2\ \times\ \bigg(\dfrac{-7}{2}\bigg)}{\dfrac{-7}{2}}\\\\\\=\ \ \ \ \dfrac{\dfrac{9}{4}\ +\ 7}{\dfrac{-7}{2}}\\\\\\=\ \ \ \ \dfrac{\bigg(\dfrac{9\ +\ 28}{4}\bigg)}{\dfrac{-7}{2}}$

$=\ \tt \ \ \ \dfrac{\dfrac{37}{4}}{\dfrac{-7}{2}}\\\\\\=\ \ \ \ \dfrac{37}{4}\ \times\ \dfrac{-2}{7}\\\\\\=\ \ \ \ \dfrac{-37}{14}$

Therefore,

$\bf \dfrac{\alpha^2\ +\ \beta^2}{\alpha\beta}\ =\ \dfrac{-37}{14}$