Question

Q1 ) Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients

(i) 5x²-29x+20
(ii) x²-5x ​

Answer 1

Step-by-step explanation:

1. 5x² - 29x + 20

= 5x² - 25x - 4x + 20

= 5x(x - 5) -4(x - 5)

= (5x - 4) (x - 5)

★ Case I

5x - 4 = 0

→ 5x = 4

$\implies \: x \: = \dfrac{4}{5}$

★ Case II

x - 5 = 0

→ x = 5

• α = $\frac{4}{5}$

• β = 5

$\large{\green{\bigstar} \: \blue{ \textsf{Sum of zeros :}}}$

$\large{\orange{ \odot} \: \pink{\alpha + \beta = \dfrac{ - b}{a} }}$

» α + β

$\dfrac{4}{5} + 5$

$= \dfrac{4 + 25}{5}$

$= \dfrac{29}{5}$

»

$\dfrac{ -b }{a}$

$= \dfrac{ - ( - 29)}{5}$

$= \dfrac{29}{5}$

Proved !

$\large{ \gray{\bigstar} \: \purple{ \textsf{Product of zeros :}}}$

$\large{ \blue{ \odot} \: \red{\alpha \beta = \dfrac{ c}{a} }}$

» αβ

$= \dfrac{4}{5} \times 5$

$= 4$

» $\dfrac{c}{a}$

$= \dfrac{20}{5}$

$= 4$

Proved !

______________________________

2. x² - 5x

= x(x - 5)

★ Case I

x = 0

★ Case II

→ x - 5 = 0

→ x = 5

• α = 0

• β = 5

$\large{\green{\bigstar} \: \blue{ \textsf{Sum of zeros :}}}$

$\large{\orange{ \odot} \: \pink{\alpha + \beta = \dfrac{ - b}{a} }}$

» α + β

= 0 + 5

= 5

Proved !

» $\dfrac{ - b}{a}$

$= \dfrac{ - ( - 5)}{1}$

= 5

$\large{ \gray{\bigstar} \: \purple{ \textsf{Product of zeros :}}}$

$\large{ \blue{ \odot} \: \red{\alpha \beta = \dfrac{ c}{a} }}$

» αβ

= 0 × 5

= 0

» $\dfrac{c}{a}$

= 0

Proved !