Question

find zeros of polynomial 4√3x^2-2√3x-2√3 and verify relations between zeros and coefficients
​

Answer 1

(a) x

2

+

6

1

x−2=0

6x

2

+x−12=0

6x

2

+9x−8x−12=0

3x(2x+3)−4(2x+3)=0

x=

3

4

,x=

2

−3

(b) 4

3

x

2

+5x−2

3

4

3

x

2

+8x−3x−2

3

=0

4x(

3

x+2)−

3

(

3

x+2)

(4x−

3

)(

3

x+2)=0

x=

4

3

,x=

3

−2

(c) 2x

2

−3x−9=0

2x

2

−6x+3x−9=0

2x(x−3)+3(x−3)=0

(2x+3)(x−3)=0

x=

2

−3

,x=3

Answer 2

Correct Question

Find zeros of polynomial 4√3x^2-5x-2√3 and verify relations between zeros and coefficients

Answer

$\tt { \color{purple}{4 \sqrt{3}x^{2} - 5x - 2 \sqrt{3} = 0 }}$

By splitting the middle terms

4√3x² - 5x - 2√3 = 0

=> 4√3x² - (8-3)x - 2√3 = 0

=> 4√3x² - 8x+ 3x - 2√3 = 0

=> 4x(√3x-2)-√3x(√3x-2) = 0

=> (√3x-2)(4x-√3x) = 0

Either,

(√3x-2) = 0

x = $\frac{2}{√3}$

Or,

(4x-√3x) = 0

x = $\frac{√3}{4}$

Therefore, the roots are $\large\frac{2}{√3}$ and $\large\frac{√3}{4}$

Verification

Sum of zeroes = $\large\rm\frac{-b}{a}$

$\mapsto\frac{2}{√3}+\frac{√3}{4}$ = $\frac{5}{4√3}$

$\mapsto\frac{5}{4√3}$ = $\frac{5}{4√3}$

Hence LHS = RHS

Product of zeroes = $\large\rm\frac{c}{a}$

$\rm\mapsto\frac{2}{√3}×\frac{√3}{4}$ = $\frac{2√3}{4√3}$

$\mapsto\frac{1}{2}$ = $\frac{1}{2}$

Here too,LHS = RHS

Hence verified!