Question

If alpha,beta are the zeroes of quadratic polynomial x2-7x+10,find the value of alpha+beta

Answer 1

Given equation

$x^{2} -7x+10=0$

Solving, we get,

$x^{2} -7x+10=0$

$\implies x^{2} -2x-5x+10=0$

$\implies x(x-2)-5(x-2)=0$

$\implies (x-5)(x-2)=0$

Now,

$x-5=0\\\implies x = 5$

$x-2=0\\\implies x=2$

Let α and β be 5 and 2 respectively

$\implies \alpha +\beta$

$\implies 5+2$

$\implies \boxed{\boxed{\bold{7}}}}$

                                                   

Answer 2

Question :

If alpha,beta are the zeroes of quadratic polynomial x^2-7x+10,find the value of alpha+beta.

Theory :

If $\sf \alpha \: and \: \beta$

are zeroes of quadratic polynomial

$\sf \: f(x) = {x}^{2} + bx + c$

Then ,$\sf \: \alpha + \beta = \dfrac{ - cofficient \: of \: x}{cofficient \: of \: x {}^{2} }$

$\sf \: and \: \alpha \beta = \dfrac{constant}{cofficient \: of \: x {}^{2} }$

Solution :

Let $\sf \: f(x) = x {}^{2} - 7x + 10$

Given alpha and beta are zeroes of f(x)

we have to find the value of alpha+ beta

$\sf \: \alpha + \beta = \frac{ - cofficient \: of \: x}{cofficient \: of \: x {}^{2} }$

$\sf \implies \: \alpha + \beta = \dfrac{ - ( - 7)}{1} = 7$