Question

If α, β are zeroes of x2+7x+7, find the value of 1/α + 1/β - 2 αβ.​

Answer 1

Step-by-step explanation:

${x}^{2} + 7x + 7 = 0 \\ \frac{1}{ \alpha } + \frac{1}{ \beta } - 2 \alpha \beta \\ = \frac{ \alpha + \beta }{ \alpha \beta } - 2 \alpha \beta \\ as \: we \: know \: that \\ \alpha + \beta = \frac{ - b}{a} \\ = > \alpha + \beta = \frac{-7}{1} \\ = > \alpha + \beta = - 7 \\ \\ \alpha \beta = \frac{c}{ \alpha } \\ = > \alpha \beta = \frac{7}{1} \\ = > \alpha \beta = 7 \\ \\ \frac{ \alpha + \beta }{ \alpha \beta } - 2 \alpha \beta \\ = \frac{-7}{7} - 2 \times 7 \\ =- 1 - 14 \\ = - 15$

Answer 2

ANSWER:

• The value of above expression = (-15)

GIVEN:

• α, β are zeroes of x²+7x+7.

TO FIND:

$\frac{1}{ \alpha} + \frac{1}{ \beta} - 2 \alpha \beta$

SOLUTION

p(x) = x²+7x+7

$\implies \: \alpha + \beta = \frac{ - (coefficient \: of \: x)}{coefficient \: of \: x {}^{2} } \\ \implies \: \alpha + \beta = \frac{ - 7}{1} \\ \\ \implies \: \alpha \beta = \frac{(constant \: term)}{coefficient \: of \: x {}^{2} } \\ \implies \alpha \beta \: = \frac{7}{1}$

Now putting the Values in the above expression.

$= (\frac{1}{ \alpha} + \frac{1}{ \beta}) - 2 \alpha \beta \\ = ( \frac{\alpha + \beta}{\alpha \beta} ) - 2\alpha \beta \\ = ( \frac{ - 7}{7} ) - (2 \times 7) \\ = - 1 - 14 \\ = - 15$

So the value of above expression = (-15)

NOTE:

some important formulas

$\implies \: \alpha + \beta = \frac{ - (coefficient \: of \: x)}{coefficient \: of \: x {}^{2} } \\ \\ \implies \: \alpha \beta = \frac{(constant \: term)}{coefficient \: of \: x {}^{2} }$