Question

5. The zeroes of the polynomial 4/3x² + 5x - 2/3 are​

Answer 1

$\sf\huge\bold{Answer:}$

Given :

$\sf:⟶ \frac{4}{3} {x}^{2} + 5x - \frac{2}{3}$

To find :

Zeroes of a given polynomial

Solution :

Step 1 : Take the polynomial = 0

$\sf:⟶ \frac{4}{3} {x}^{2} + 5x - \frac{2}{3} =0$

Step 2 : Taking LCM

$\sf:⟶ \frac{4 {x}^{2} + 15x - 2 }{3} = 0$

Step 3 : Taking 3 to opposite side to simplify equation

$\sf:⟶ 4 {x}^{2} + 15x - 2 = 0 × 3$

$\sf:⟶ 4 {x}^{2} + 15x - 2 = 0$

Step 4 : Using discriminant method to find zeroes

① Compare given equation with general equation to get values of a, b, c

$\sf:⟶ 4 {x}^{2} + 15x - 2 = 0$

$\sf:⟶ a {x}^{2} + bx + c = 0$

a=4

b=15

c= -2

② Find discriminant

$\fbox\purple{Formula used}$

:⟶ D=b²-4ac

:⟶D=15²-4×4×(-2)

:⟶D=225-16×(-2)

:⟶D=225-(-32)

:⟶D=225+32

:⟶D=257

③ To find zeroes

$\fbox\purple{Formula Used}$

$\sf:⟶x = \frac{ - b± \sqrt{D} }{2a}$

$\sf:⟶x = \frac{ - 15 ± \sqrt{257} }{2×4}$

$\sf:⟶x = \frac{ - 15 ± \sqrt{257} }{8}$

[NOTE] : 257 is a prime number, so it will remain as √257 in the equation

Hence, zeroes are

$\sf\huge:⟶x = \frac{ - 15 ± \sqrt{257} }{8}$

i.e.

$\huge\purple{:⟶x = \frac{ - 15 + \sqrt{257} }{8}}$ and

$\huge\purple{:⟶x = \frac{ - 15 - \sqrt{257} }{8}}$