Question

Find the zeros of quadratic polynomial X ^ 2 - 5x +6 between the zeros and the coefficients

Answer 1

Given :

• Quadratic Polynomial x²-5x+6.

To Find :

• Zeroes of the polynomial.

Solution :

$\longmapsto\tt\bold{{x}^{2}-5x+6}$

By Splitting Middle Term :-

$\longmapsto\tt{{x}^{2}-(3x+2x)+6}$

$\longmapsto\tt{{x}^{2}-3x-2x+6}$

$\longmapsto\tt{x(x-3)-2(x-3)}$

$\longmapsto\tt{(x-2)(x-3)}$

• x = 2
• x = 3

So , 2 and 3 are the zeroes of polynomial x²-5x+6..

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Here :

• a = 1
• b = -5
• c = 6

Sum of Zeroes :

$\longmapsto\tt{\alpha+\beta=\dfrac{-b}{a}}$

$\longmapsto\tt{2+3=\dfrac{-(-5)}{1}}$

$\longmapsto\tt\bold{5=5}$

Product of Zeroes :

$\longmapsto\tt{\alpha\beta=\dfrac{c}{a}}$

$\longmapsto\tt{2\times{3}=\dfrac{6}{1}}$

$\longmapsto\tt\bold{6=6}$

HENCE VERIFIED

Answer 2

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$\huge{\mathcal{\underline{\red{SolutioN}}}}$

$\longmapsto\tt\bold{{x}^{2}-5x+6}</p><p> \:$

⍟$\tt \: By \: spilting \: the \: middle \: term$

$\longmapsto\tt{{x}^{2}-(3x+2x)+6} \\ </p><p>\longmapsto\tt{{x}^{2}-3x-2x+6} \\ </p><p>\longmapsto\tt{x(x-3)-2(x-3)} \\ </p><p>\longmapsto\tt{(x-2)(x-3)}$

• x = 2
• x = 3

$\tt so .. \\ \tt 2 \: and \: 3 \: are \: the \tt \: zeros \: of \: the \: polynomial \: \\ \tt{x}^{2} - 5x + 6 \:$

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⍟ $\tt{Here}$

• a = 1
• b = -5
• c = 6

⍟ $\tt Sum \: of \: Zeros \:$

$\longmapsto\tt{\alpha+\beta=\dfrac{-b}{a}} \\ </p><p>\longmapsto\tt{2+3=\dfrac{-(-5)}{1}} \\ </p><p>\longmapsto\tt\bold{5=5}$

⍟$\tt \: Product \: of \: Zeros \:$

$\longmapsto\tt{\alpha\beta=\dfrac{c}{a}} \\ </p><p>\longmapsto\tt{2\times{3}=\dfrac{6}{1}} \\ </p><p>\longmapsto\tt\bold{6=6}$

⍟$\tt\red \: Hence \: Proofed$

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