Question

Find the zeroes of the quadratic polynomial x2 + 7x +10 and verify the relation between the zeroes

and coefficients

by by​

Answer 1

This is the answer....

Answer 2

Answer:

Question :-

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

Required Answer :-

Solution :-

p(x) = x² + 7x + 10

⇒General Form :-

ax² + bx + c

⇒Given Quadratic polynomial :-

x² + 7x + 10

so here we get a value of a, b and c

a = 1

b = 7

c = 10

⇒Let's find the zeroes by splitting the middle term

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

$= {x}^{2} + 2x + 5x + 10$

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

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

Hence the value of x² + 7x + 10 is zero then either x+5 = 0 or x+2 = 0, i.e, when x = -5 or x = -2.

x + 5 = 0⠀⠀⠀⠀⠀⠀⠀⠀⠀x + 2 = 0

x = - 5⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀x = - 2

So, the zeroes of x² + 7x + 10 are x - 5 and x + 5.

⇒Verifying the relationship between zeroes and coefficients

let one zero be alpha and other be beta

⇒Sum of zeroes :-

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

$- 5 + ( - 2) = \frac{ - 7}{1}$

$- 5 - 2 = - 7$

$- 7 = - 7$

⇒Product of zeroes :-

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

$- 5 \times - 2 = \frac{10}{1}$

$10 = 10$

Hence verified