Question

find the zeros of the quadratic polynomial 2 x square - 7 x + 5 is equals to zero and verify the relationship between zeros and coefficients​

Answer 1

Answer:

Zeroes are (5/2) and (1)

Step-by-step explanation:

Given : f(x) = 2x² - 7x + 5

By Middle Term Factorisation

→ 2x² - 2x - 5x + 5

Taking common terms out.

→ 2x(x - 1) - 5(x - 1)

→ (2x - 5)(x - 1)

To find the zeroes, we use zero product rule.

→ (2x - 5) = 0 and (x - 1) = 0

→ x = 5/2 and x = 1

Let α and β be the zeroes of the above polynomial.

∴ α = 5/2 & β = 1

______________________________

On comparing the above polynomial with ax² + bx + c, we get

a = 2, b = - 7, c = 5

______________________________

Verification:

• Sum of zeroes = α + β

→ 5/2 + 1

→ (5 + 2)/2

→ 7/2

Also, Sum of zeroes = - b/a

→ - (- 7)/2

→ 7/2

• Product of zeroes = αβ

→ (5/2)(1)

→ 5/2

Also, Product of zeroes = c/a

→ 5/2

______________________________

Answer 2

AnswEr :

$\underline{\bigstar\:\textsf{According \: to \: given \: in \: question:}}$

$\normalsize\ : \implies\sf\ P(x) = 2x^2 - 7x + 5$

$\scriptsize\sf{\: \: \: \: \: \:( \therefore\ \: \red{Using \: middle \: term \: factorization}) }$

$\normalsize\ : \implies\sf\ 2x^2 - 2x - 5x +5 \\ \\ \normalsize\ : \implies\sf\ 2x(x - 1) - 5(x - 1) \\ \\ \normalsize\ : \implies\sf\underbrace{(x - 1)}_{case \: 1} \: \underbrace{(2x - 5)}_{case \: 2}$

$\rule{100}2$

$\:\bullet\:\sf\ Case \: 1 :$

$\normalsize\ : \implies\sf\ (x - 1) = 0 \\ \\ \normalsize\ : \implies\sf\ x = 0 + 1 = 1$

$\normalsize\ : \implies{\boxed{\sf \green{ x = 1}}}$

$\:\bullet\:\sf\ Case \: 2:$

$\normalsize\ : \implies\sf\ (2x - 5) = 0 \\ \\ \normalsize\ : \implies\sf\ x = \frac{5}{2}$

$\normalsize\ : \implies{\boxed{\sf \green{ x = \frac{5}{2} }}}$

$\:\bullet$ Here, we get x = 1, 5/2, let the one zero be $\bf\alpha$ and other be $\bf\beta$.So;

$\:\star\normalsize\sf\:\alpha = 1 \\ \\ \:\star\normalsize\sf\:\beta = \frac{5}{2}$

$\:\bullet$ Now; compare the given polynomial with general form of polynomial $\normalsize\sf\ [ax^2 + bx + c]$, we get a = 2, b = -7, c = 5.

$\rule{100}2$

VeriFicatioN :

$\:\bullet$ $\normalsize\frak{\underline{\underline \purple{Sum \: of \: roots-}}}$

$\normalsize\ : \implies\sf\ \alpha + \beta = \frac{-b}{a} \\ \\ \normalsize\ : \implies\sf\ 1 + \frac{5}{2} = \frac{-(7)}{2} \\ \\ \normalsize\ : \implies\sf\ \frac{7}{2} = \frac{7}{2} \\ \\ \normalsize\ : \implies\sf\ L.H.S = R.H.S$

$\:\bullet$ $\normalsize\frak{\underline{\underline \purple{Product \: of \: roots-}}}$

$\normalsize\ : \implies\sf\ \alpha\beta = \frac{c}{a} \\ \\ \normalsize\ : \implies\sf\ 1 \times\ \frac{5}{2} = \frac{5}{2} \\ \\ \normalsize\ : \implies\sf\ L.H.S = R.H.S$

$\normalsize\ : \implies\sf\ Hence \: VeriFieD \: !!$