Question

Find the zeroes of the quadratic polynomials i) x2 + 5x + 6 ii) x2 –(√3 + 1)x + √3 and

verify the relationship between the zeroes and their coefficients.​

Answer 1

Question :

Find the zeroes of the quadratic polynomials -

i) x² + 5x + 6

ii) x² –(√3 + 1)x + √3

and verify the relationship between the zeroes and their coefficients.

Solution :

(1)

f(x) = x² + 5x + 6

> f(x) = x² + 2x + 3x + 6

> f(x) = x( x + 2) + 3(x+2)

> f(x) = (x+3)(x+2)

The zeroes are -2 and -3 respectively .

Sum of zeroes = -5 = -b/a = (-5/1) = -5

Product of zeroes = 6 = c/a = 6/1 = 6 .

Hence Verified

(2)

f(x) = x² - (√3+1)x + √3

> f(x) = x² - √3x - x + √3

> f(x) = x(x-√3) -(x-√3)

> f(x) = (x-1)(x-√3)

The zeroes are 1 and √3 respectively .

Sum of zeroes = ( √3+1) = -b/a -(√3+1)/1 = -(√3+1)

Product of zeroes = √3 = c/a = √3/1 = √3

Hence Verified

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Answer 2

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G I V E N :

• Quadratic polynomials ℹ) x² + 5x + 6 and ℹℹ) x² - (√3 + 1)x + √3.

T OㅤF I N D :

• Zeroes of both polynomials.
• Also we have to verify relationship b/w their zeroes and coefficients.

S O L U T I O N :$\:$

✇ Finding zeroes of 1st polynomial :-

➵ㅤㅤㅤx² + 5x + 6 = 0

✇ Splitting the middle term :-

➵ㅤㅤㅤx² + 2x + 3x + 6 = 0

➵ㅤㅤㅤx(x + 2) + 3(x + 2) = 0

➵ㅤㅤㅤ(x + 2) (x + 3) = 0

➵ㅤㅤㅤx + 2 = 0 , x + 3 = 0

➵ㅤㅤㅤx = -2 , x = -3

∴ Hence, zeroes of polynomial x² + 5x + ㅤ6 (α and β) = -2 and -3.

✇ Verifying relationship b/w it's zeroes ㅤand coefficients :-

➵ㅤα + β = -b/a , αβ = c/a

Where,

• α and β = -2 and -3
• b = coefficient of x = 5
• c = constant term = 6
• a = coefficient of x² = 1

✇ Substituting all known values :-

➵ㅤ-2 + (-3) = -(5)/1 , (-2)(-3) = 6/1

➵ㅤ-2 - 3 = -5 , 2 × 3 = 6

➵ㅤ-5 = -5 , 6 = 6

ㅤㅤㅤㅤㅤ∴ Hence, Verified!

✇ Finding zeroes of 2nd polynomial :-

➵ㅤㅤㅤx² - (√3 + 1)x + √3 = 0

➵ㅤㅤㅤx² - (√3x + x) + √3 = 0

➵ㅤㅤㅤx² - √3x - x + √3 = 0

➵ㅤㅤㅤx(x - √3) - 1(x - √3) = 0

➵ㅤㅤㅤ(x - √3) (x - 1) = 0

➵ㅤㅤㅤx - √3 = 0 , x - 1 = 0

➵ㅤㅤㅤx = √3 , x = 1

∴ Hence, zeroes of polynomial x² - (√3 + ㅤ1)x + √3 (α and β) = √3 and 1.

✇ Verifying relationship b/w it's zeroes ㅤand coefficients :-

➵ㅤα + β = -b/a , αβ = c/a

Where,

• α and β = √3 and 1
• b = coefficient of x = 5
• c = constant term = 6
• a = coefficient of x² = 1

✇ Substituting all known values :-

➵ㅤ√3 + 1 = -[-(√3 + 1)]/1 , √3 × 1 = √3/1

➵ㅤ(√3 + 1) = (√3 + 1) , √3 = √3

ㅤㅤㅤㅤㅤ∴ Hence, Verified!

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