Question

Find the zeros of quadratic polynomial x²+9x+14 and verify the relation between zeros and coefficients

Answer 1

Given:-

A quadratic polynomial x²+9x+14 .

To find out:-

Find the zeros of quadratic polynomial and verify the relation between zeros and coefficients.

Solution:-

We have,

x²+9x+14

= x² + 2x + 7x + 14

= x ( x + 2 ) + 7 ( x + 2 )

= ( x + 2 ) ( x + 7 )

= x + 2 = 0 and ( x + 7 ) = 0

= x = -2 and x = -7

Therefore,the zeroes of x²+9x+14 are -2 and -7 .

Verification:-

★ Sum of the zeroes = -coefficient of x/coefficient of x²

⇒ -2 + ( - 7 ) = - 9 / 1

⇒ -9 = -9

L.H.S = R.H.S

★ Product of the zeroes = constant term/coefficient of x²

⇒ -2 × -7 = 14/1

⇒ 14 = 14

L.H.S = R.H.S

Verified.

Answer 2

Heya!!

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Given,

Quadratic polynomial = ${X}^{2}$ +9x+14

To Find:-

➡️Zeros of the polynomial

➡️ Verification between the zeros

➡️ Verification between coefficients.

Calculation:-

${x}^{2} + 9x + 14$

$= {x}^{2} + 2x + 7x + 14$

$= x(x + 2) + 7(x + 2)$

$= (x + 2)(x + 7)$

$= x + 2 = 0$

$= x + 7 = 0$

so,

$x = - 2and \: - 7$

The zeros of $x^{2}$ is -2 and -7.

$\underline{\sf{Verification}}$

Sum of zeroes = $\frac{-coefficient\:of\:X}{coefficient \:of\:X^{2}}$

➡️-2+(-7) = $\frac{-9}{1}$

➡️-9 = -9

LHS = RHS

Product of zeroes = $\frac{Constant\:term}{coefficient\:of\:x^{2}}$

= -2×(-7) = 14/1

= 14 = 14

LHS = RHS.

Explore more:-

We known that

- × - = +

+ × + = +

- × + = -

+ × - = -