Question

14.Find the zeroes of the quadratic polynomial x2+x-12 and verify the relationship between the zeroes and the coefficients

Answer 1

$\huge\mathfrak{Answer:}$

Given:

• We have been given a polynomial x² + x - 12.

To Find:

• We need to find the zeroes of this polynomial.

Solution:

The given polynomial is x² + x - 12.

We can find the zeroes of this quadratic polynomial by the method of splitting the middle term.

We need to find two such numbers whose sum is 1 and product is -12.

Two such numbers are 4 and -3.

Substituting the values, we have

x² + 4x - 3x - 12 = 0

=> x(x + 4) - 3(x + 4) = 0

=> (x + 4) (x - 3) = 0

Either (x + 4) = 0 or (x - 3) = 0.

When (x + 4) = 0

=> x = -4

When (x - 3) = 0

=> x = 3

Hence, two zeroes of this polynomial are -4 and 3.

Now, we need to verify the relationship between the zeroes and coefficients, we have

A = -4

B = 3

Sum of zeroes ( A + B )

= -4 + 3

= -1

= -b/a _______(1)

Product of zeroes ( AB )

= -4 × 3

= -12

= c/a ________(2)

From equation 1 and 2 relationship between zeroes and coefficients is verified!!

Answer 2

✯✯ QUESTION ✯✯

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

━━━━━━━━━━━━━━━━━━━━

✰✰ ANSWER ✰✰

$\implies\tt{{x}^{2}+x-12=0}$

➮By Splitting Middle Term : -

$\implies\tt{{x}^{2}+(4x-3x)-12=0}$

$\implies\tt{{x}^{2}+4x-12=0}$

$\implies\tt{x(x+4)-3(x+4)=0}$

$\implies\tt{(x-3)(x+4)=0}$

• x = 3
• x = -4

☛So , 3 and -4 are the zeroes of polynomial of x²+x-12..

_______________________

➥VERIFICATION : -

➮Here : -

• a = 1
• b = 1
• c = -12

★Sum of Zeroes : -

$\implies\tt{3+(-4)=\dfrac{c}{a}}$

$\implies\tt{3-4=\dfrac{-1}{1}}$

$\implies\tt{-1=-1}$

$\red\longmapsto\:\large\underline{\boxed{\bf\green{L.H.S}\orange{=}\purple{R.H.S}}}$

★Product of Zeroes : -

$\implies\tt{3\times{-4}=\dfrac{-12}{1}}$

$\implies\tt{-12=-12}$

$\red\longmapsto\:\large\underline{\boxed{\bf\orange{L.H.S}\green{=}\pink{R.H.S}}}$

✺ HENCE VERIFIED ✺