Question

Find the zeroes of 2√3x^2-5x+√3
verify the relationship between zeroes and coefficient
,​

Answer 1

Answer:-

$\sf{Its \: zeros \: are \: \to \: \frac{1}{ \sqrt{3} } \: \: and \: \frac{ \sqrt{3} }{2} } \\ \:$

To Find :-

→ Zeros and relationships between coefficient and zeros .

Explanation :-

We have ,

$\to \sf{2 \sqrt{3} {x}^{2} - 5x + \sqrt{3} = 0}$

split the middle term

$\to \sf{ \: 2 \sqrt{3} {x}^{2} - (3 + 2)x + \sqrt{3} = 0} \\ \\ \to \sf{2 \sqrt{3} {x}^{2} - 3x - 2x + \sqrt{3} = 0} \\ \\ \to \sf{ 2 \sqrt{3}{x}^{2} - \sqrt{3} \sqrt{3} x - 2x + \sqrt{3} = 0} \\ \\ \to \sf{\sqrt{3} x(2x - \sqrt{3} ) - 1(2x - \sqrt{3} ) = 0} \\ \\ \to \sf{ (2x - \sqrt{3} )( \sqrt{3}x - 1) = 0} \\ \\ \bf{\underline{case \: (1)} \: if \: } \\ \\ \to \sf{\sqrt{3} x - 1 = 0} \\ \\ \to \sf{\sqrt{3} x = 1} \\ \\ \to \boxed{\sf{ x = \frac{1}{ \sqrt{3} } }}$

$\bf{\underline{case \: (2)} \: if \: } \\ \to \sf{ 2x - \sqrt{3} = 0} \\ \\ \to \sf{2x = \sqrt{3} } \\ \\ \to \boxed{\sf{x = \frac{ \sqrt{3} }{2} }}$

$\sf{Its \: zeros \: are \: \to \: \frac{1}{ \sqrt{3} } \: \: and \: \frac{ \sqrt{3} }{2} }$

Relation between zeros and coefficients :-

Here ,

$\to \sf{sum \: of \: zeros = \frac{ - b}{a} = \frac{5}{2 \sqrt{3} } } \\ \\ \: \: \: \: \frac{1}{ \sqrt{3} } + \frac{ \sqrt{3} }{2} = \frac{5}{2 \sqrt{3} } \\ \\ \: \: \: \: \frac{2 + 3}{2 \sqrt{3} } = \frac{5}{ 2\sqrt{3} } \\ \\ \: \: \boxed{ \: \: \frac{5}{2 \sqrt{3} } = \frac{5}{2 \sqrt{3} } }$

and ,

$\sf{product \: of \: zeros = \frac{c}{a} = \frac{ \sqrt{3} }{2 \sqrt{3} } } \\ \\ \: \: \: \: \: \frac{1}{ \sqrt{3} } \times \frac{ \sqrt{3} }{2} = \frac{ \sqrt{3} }{2 \sqrt{3} } \\ \\ \: \: \: \: \: \boxed{ \frac{1}{2} = \frac{1}{2} }$

hence verified .

Answer 2

Answer:-

Given polynomial => 2√3x² - 5x + √3 = 0

By splitting the middle term,

2√3x² - 2x - 3x + √3 = 0

2x(√3x - 1) - √3(√3x - 1) = 0

(2x - √3)(√3x - 1) = 0

2x - √3 = 0

2x = √3

x = √3/2

√3x - 1 = 0

√3x = 1

x = 1/√3

Verification:-

Let , a = 2√3 ; b = - 5 and c = √3

We know that,

Sum of the roots = - b/a

√3/2 + 1/√3 = -(-5)/2√3

By taking LCM,

(3 + 2)/2√3 = 5/2√3

5/2√3 = 5/2√3

Product of zeroes = c/a

(√3/2)(1/√3) = √3/2√3

1/2 = 1/2

Hence, Verified.