Question

Find the zeros of the given polynomials and verify the relationship between the zeros and their coefficients x square - 10x + 25

Answer 1

Given:-

→p(x)=0

→x²-10x+25=0

$\rule{200}{1}$

To Find:-

→The Zeros of P(x) and verify the relationship b/w their coefficients?

$\rule{200}{1}$

AnsWer:-

→x²-10x+25=0

๛S=-10[Sum of Equation]

๛P=25[Product of Equation]

•-5-5=-10=S

•-5×-5=25=P

★Since there are common factors for p(x),We will use Spiliting middle term★

→x²-5x-5x+25=0

→x(x-5)-5(x-5)=0

→x-5=0

→x=5 are the roots

๛α=5

๛β=5

$\rule{200}{1}$

♦Verification:-

→α+β=$\frac{-b}{a}$

→5+5=$\frac{-(-10)}{1}$

→10=10

★LHS=RHS★

✓erified

$\rule{200}{1}$

♦Verification:-

→αβ=$\frac{c}{a}$

→5×5=$\frac{25}{1}$

→25=25

★LHS=RHS★

✓erified

$\rule{200}{2}$

Answer 2

Answer:

⋆ Given Polynomial : x² - 10x + 25

Here : a = 1,⠀b = - 10,⠀c = 25

$:\implies\tt f(x) = 0\\\\\\:\implies\tt x^2 - 10x + 25= 0\\\\\\:\implies\tt (x)^2 - (2 \times x \times 5) +(5)^2= 0\\\\{\scriptsize\qquad\bf{\dag}\:\:\sf{(a)^2-2ab+(b)^2=(a-b)^2}}\\\\:\implies\tt (x - 5)^2 = 0\\\\\\:\implies\tt x - 5=0\\\\\\:\implies\underline{\boxed{\tt x =5}}$

$\rule{150}{1}$

$\underline{\bigstar\:\:\textsf{Relation b/w zeroes and coefficient :}}$

$\qquad\underline{\bf{\dag}\:\:\textsf{Sum of Zeroes :}}\\\dashrightarrow\tt\:\: \alpha+\beta = \dfrac{-\:b}{a}\\\\\\\dashrightarrow\tt\:\: 5+5 = \frac{ -\:(-10)}{1}\\\\\\\dashrightarrow\:\:\underline{\boxed{\red{\tt 10=10}}} \\\\\\\qquad\underline{\bf{\dag}\:\:\textsf{Product of Zeroes :}}\\\\\dashrightarrow\tt\:\: \alpha \times \beta = \dfrac{c}{a}\\\\\\\dashrightarrow\tt\:\: 5\times5 = \dfrac{25}{1}\\\\\\\dashrightarrow\:\:\underline{\boxed{ \red{\tt 25 = 25}}}$