Question

If α and β are the zeroes of f(x) = 2x2 + 8x – 8, then prove that α + β – αβ = 0

Answer 1

Answer:

$\alpha \: \: \: \: \beta \: are \: the \: zeros \: of \: the \: equation \: \: \\ 2 {x}^{2} + 8x - 8 \\ \\ \alpha + \beta = \frac{ - 8}{2} = - 4 \\ \\ \alpha \beta = \frac{ - 8}{2} = - 4 \\ \\ now \: \: l \: h \: s = \: \: \alpha + \beta - \alpha \beta \\ = - 4 - ( - 4) \\ = - 4 + 4 \\ = 0 \\ = r \: h \: s \\ \: proved$

Answer 2

✤Qᴜᴇꜱᴛɪᴏɴ :-

If α and β are the zeroes of f(x) = 2x² + 8x – 8, then prove that α + β – αβ = 0

✤ꜱᴏʟᴜᴛɪᴏɴ :-

$\bf f(x) = 2x² + 8x – 8$

Value of α + β

$\bf \to \alpha + \beta = \frac{ - b}{a}$

$\bf \to \alpha + \beta = \frac{ - 8}{2}$

$\bf \to \alpha + \beta = - 4⠀⠀⠀ eq.1$

Value of αβ

$\bf \to \alpha \beta = \frac{c}{a}$

$\bf \to \alpha \beta = \frac{ - 8}{2}$

$\bf \to \alpha \beta = - 4⠀⠀⠀eq.2$

Subtracting equation 1 from equation 2

$\bf \implies \alpha + \beta - \alpha\beta = 0$

$\bf \implies - 4 - ( - 4) = 0$

$\bf \implies - 4 + 4 = 0$

$\bf \implies 0 = 0$

$\bf \implies L.H.S = R.H.S$

$\bf \pink{Hence, } \purple{proved}$