Question

12
Find the zeroes of the quadratic polynomial x²+7x + 10. and verify the relationship between the
zeroes and the coefficients.​

Answer 1

$\huge\underline{\bf \orange{Solution :}}$

By splitting middle term

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

let's α and β are the zeroes of the polynomial , So

• α = - 2
• β = - 5

$\huge\underline{\bf \orange{Verification :}}$

$\implies \boxed{\sf \alpha + \beta = \frac{ - \: Coefficient \: of \: x}{Coefficient \: of \: {x}^{2}} } \\ \\ \implies \sf - 2 + ( - 5) = \frac{ - 7}{1} \\ \\ \implies \sf - 2 - 5= - 7 \\ \\ \implies \sf - 7= - 7$

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$\implies \boxed{\sf \alpha \times \beta = \frac{ Constant \:term}{Coefficient \: of \: {x}^{2}} } \\ \\ \implies \sf (- 2 )\times ( - 5) = \frac{10}{1} \\ \\ \implies \sf 10 = 10$