Question

find the zeros of the polynomial p(x)=x^2-4x+3 and verify the relationship between zeroes and coefficient​

Answer 1

${\bold{\underline{\underline{Answer:}}}}$

${\bold{\therefore x=3\:and\:1}}$

${\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\underline \bold{given : }\\ \implies {x}^{2} - 4x + 3 = 0 \\ \\ \underline \bold{to \: find : } \\ \implies x = ?$

• According to given question :

$\bold{First \: method : } \\ \\ \bold{Middle \: term \: spliting : } \\ \implies {x}^{2} - 4x + 3 = 0 \\ \\ \implies {x}^{2} - 3x - x + 3 = 0 \\ \\ \implies x(x - 3) - 1(x - 3) = 0 \\ \\ \implies (x - 3)(x - 1) = 0 \\ \\ \bold{\implies x = 3 \: and \: 1} \\ \\ \bold{Second \: method : } \\ \\ \bold{Quadratic \: formula : } \\ \implies x = \frac{ - b \pm \sqrt{ {b}^{2} - 4ac} }{2a} \\ \\ \implies x = \frac{ - ( - 4) \pm \sqrt{( { - 4)}^{2} - 4 \times 3 } }{2 \times 1} \\ \\ \implies x = \frac{4 \pm \sqrt{16 - 12} }{2} \\ \\ \implies x = \frac{4 \pm 2}{2} \\ \\ \bold{\implies x = 3 \: and \: 1}\\\\\bold{Sum\:of\:zeroes:}\\\implies3+1\\\bold{\implies 4}\\\\\bold{Relationship:}\\\implies \frac{-Coefficient\:of\:x}{Coefficient\:of\:x^{2}}\\\\\implies \frac{-(-4)}{1}\\\\\bold{\implies 4}\\\\{\bold{\boxed{Verified}}}$

Answer 2

Answer:

Hey mate

I will help

Step-by-step explanation:

$p(x) = {x}^{2} - 4x + 3$

Hence

$p(x) = 0$

We can solve this by using splitting middle term

Therefore,

${x}^{2} - 3x - x + 3 = 0$

Taking common

$x(x - 3) -1 (x - 3) = 0$

$(x - 1) = 0 \: or \: (x - 3) = 0$

$x = 1 \: and \: x = 3$

Therefore,

$\alpha = 1 \: and \: \beta = 3$

Verify the relationship,

$\alpha + \beta = 1 + 3 \\ \: \: \: \: \: \: \: \: \: \: \: = 4 \\ also \\ - \frac{b}{a} = \frac{ - ( - 4)}{?1} = + 4$

$therefore \: \\ \: \alpha + \beta = \frac{ - b}{a}$

$\alpha \beta = 1 \times 3 = 3 \\ \frac{c}{a} = \frac{3}{1} = 3$

$\alpha \beta = \frac{c}{a}$

Hence solve