Question

1. Authan maths:
0
Find the sum of the zeroes of polynomial x²-6x+7​

Answer 1

GIVEN :-

• p(x) = x² - 6x + 7

TO FIND :-

• The sum of zeroes of the given polynomial

SOLUTION :-

Let the roots of the given quadratic equation be α , β . Then ;

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

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

where ,

• a is coefficient of x²
• b is coefficient of x
• c is constant

Now , From the given polynomial ;

• a = 1
• b = -6
• c = 7

First let us find the value of α. Now , By substituting the values we get ;

$\\ : \implies \sf \alpha \: = \frac{ - ( - 6) + \sqrt{( - 6) {}^{2} - 4(1)(7) } }{2(1)}$

$\\ : \implies \sf \alpha = \frac{6 + \sqrt{36 - 28} }{2}$

$\\ : \implies \sf \alpha = \dfrac{6 + \sqrt{8} }{2}$

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

$\\ : \implies \sf \alpha = \dfrac{ 2(3 + \sqrt{2})}{2}$

$\\ : \implies \sf \alpha \: = 3 + \sqrt{2}$

• Now let us Find the value of β ,

$\\ : \implies \sf \beta = \frac{ - ( - 6) - \sqrt{( - 6) {}^{2} - 4(1)(7) } }{2(1)}$

$\\ : \implies \sf \beta = \frac{6 - \sqrt{36 - 28} }{2}$

$\\ : \implies \sf \beta = \frac{6 - \sqrt{8} }{2}$

$\\ :\implies\sf\beta=\dfrac{6-2\sqrt{2}}{2}$

$\\ : \implies \sf \beta = \frac{2(3 - \sqrt{2} )}{2}$

$\\ : \implies \sf \beta = 3 - \sqrt{2}$

Now we need to find the sum of zeroes . sum of zeroes is nothing but the addition of α and β.

$\\ : \implies \sf \alpha + \beta = (3 + \sqrt{2} ) + (3 - \sqrt{2} )$

$\\ : \implies \sf \alpha \: + \beta \: = 6 + 0$

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

Hence , The sum of the zeroes of x² - 6x + 7 is 6.

Answer 2

Answer:

6 . Sum of the roots of x^2-6x+7 is 6