Question

find the sum and the product of the zeroes of the polynomial x2-6x+5​

Answer 1

Step-by-step explanation:

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Answer 2

GIVEN :-

• A quadratic equation x² - 6x + 5 = 0.

TO FIND :-

• sum of zeros ( ɑ + β )
• product of zeros ( ɑ × β ).

SOLUTION :-

$\\ : \implies \displaystyle \sf \: \alpha = \frac{ - b + \sqrt{b ^{2} - 4ac } }{2a}$

• a = 1.
• b = -6.
• c = 5.

$\\ \\ : \implies \displaystyle \sf \: \alpha = \frac{ - ( - 6) + \sqrt{( - 6) ^{2} - 4 \times 1 \times 5 } }{2 \times 1}$

$: \implies \displaystyle \sf \: \alpha = \frac{6 + \sqrt{36 - 20} }{2}$

$: \implies \displaystyle \sf \: \alpha = \frac{6 + \sqrt{16} }{2}$

$: \implies \displaystyle \sf \: \alpha = \frac{6 + 4}{2}$

$: \implies \displaystyle \sf \: \alpha = \frac{10}{2}$

$: \implies \underline{ \boxed{ \displaystyle \sf \: \alpha = 5}}$

Similarly,

$\\ \\ : \implies \displaystyle \sf \: \beta = \frac{ - b - \sqrt{b ^{2} - 4ac } }{2a}$

• a = 1.
• b = -6.
• c = 5.

$\\ \\ : \implies \displaystyle \sf \: \beta = \frac{ - ( - 6) - \sqrt{( - 6) ^{2} - 4 \times 1 \times 5 } }{2 \times 1}$

$: \implies \displaystyle \sf \: \beta = \frac{6 - \sqrt{36 - 20} }{2}$

$: \implies \displaystyle \sf \: \beta = \frac{6 - \sqrt{16} }{2}$

$: \implies \displaystyle \sf \: \beta = \frac{6 - 4}{2}$

$: \implies \displaystyle \sf \: \beta = \frac{2}{2}$

$: \implies \underline{ \boxed{\displaystyle \sf \: \beta = 1}}$

Now sum of zeroes ( ɑ + β ),

$\\ : \implies \displaystyle \sf \: \alpha + \beta = 5 + 1$

$: \implies \underline{ \boxed{\displaystyle \sf \: \alpha + \beta = 6}}$

Similarly product of zeroes ( ɑ × β ),

$\\ : \implies \displaystyle \sf \: \alpha \beta = 5 \times 1$

$: \implies \underline{\boxed{ \displaystyle \sf \: \alpha \beta = 5}}$