Question

Find the zeroes of the quadnatic
polynomial x² + 7 x + 10 and verify the
relationship between the zeroes and
the cofficients​

Answer 1

$\huge{\underline{\underline{\red{Solution→}}}}$

${x}^{2} + 7x + 10$

${x}^{2} + 5x + 2x + 10 \\ {x}(x + 5) + 2(x + 10) \\ (x + 2)(x + 5)$

Zeros of this Equation are following :-

$x + 2 = 0 \\ x = - 2$

$\rule{200}{1}$

$x + 5 = 0 \\ x = - 5$

$\rule{200}{1}$

The relationship between Zeros and Coefficient :-

• $sum \: of \: zeros \: = - \frac{coefficient \: of \: x}{coefficient \: of \: {x}^{2} }$

• $product \: of \: zeros \: = \frac{constant \: term}{coefficient \: of \: {x}^{2} }$

$\rule{200}{1}$

In the Given Polynomial :-

• The Coefficient of x² = 1

• The Coefficient of x = 7

• The constant term = 10

$\rule{200}{1}$

So, the relationship →

$sum \: of \: zeros \: = - \frac{7}{1} \\ - 2 - 7 = - 7 \\ - 7 = - 7$

LHS = RHS

$\rule{200}{1}$

$product \: of \: zeros \: = \frac{10}{1} \\ - 2 \times (- 5) = 10 \\ 10 = 10$

LHS = RHS

Hence Proved !!

$\rule{200}{1}$

Answer 2

Answer:

Step-by-step explanation:

x^2 + 7x + 10

= x^2 + 5x + 2x + 10

= x(x + 5) + 2(x + 5)

= (x + 5)(x + 2)

α = -5, β = -2

Verification:

α + β = -b/a

-5 + (-2) = -7/1

-7 = -7

αβ = c/a

(-5)(-2) = 10/1

10 = 10

Hence verified.

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