Question

D 09 8. Find the zeroes of the quadratic polynomial x2 + 7x + 10 and verify the relationship between the zeroes and the coefficients.​

Answer 1

${x}^{2} + 7x + 10 \\ {x}^{2} + 5x + 2x + 10 \\ x(x + 5) \: \: \: \: 2(x + 5) \\( x + 2) \: \:( x + 5) \\( x + 2) = 0 \\ x = - 2$

$x + 5 = 0 \\ x = - 5 \\ \alpha = - 2 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \beta = - 5 \: \: are \: the \: zeroes \: of \: the \\ polynomial$

$\alpha = - 2 \: \: \: \: \: \: \: \: \: \beta = - 5 \\ \alpha + \beta = - 2 + - 5 = - 7 \\ \alpha \beta = - 2 \times - 5 = 10 \\ \\ {x}^{2} + 7x + 10 \\ \: \: \: \: \: \:$

$a = 1 \: \: \: \: \: \: \: \: b = 7 \: \: \: \: \: \: \: \: c = 10 \\ \frac{ - b}{a} = \frac{ - coeffient \: of \: x}{coefficient \: of \: {x}^{2} } = \frac{-7}{1} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = - 7 \\ \frac{c}{a} = \frac{constant \: term}{coefficient \: of \: {x}^{2} }=\frac{10}{1}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 10 \\ \alpha + \beta = \frac{ - b}{a} = - 7 \\ \alpha \beta = \frac{c}{a} = 10 \\ hence \: verified$