Question

Find the zeros of quadratic Polynomial 4√3x^2+5x-2√3 .Also verify the relationship between zeros and its coefficients​

Answer 1

Given Polynomial → 4√3x² + 5x - 2√3

For finding zeros of any polynomial first, factorize the given polynomial by splitting the middle term.

So, for this find 2 numbers such that

• Their product is (4√3)(5√3) = 24
• Their sum is 5

Those 2 numbers which satisfy the above condition are 8 and 3.

   4√3x² + 5x - 2√3 = 0

⇒ 4√3x² + 8x - 3x - 2√3 = 0

⇒ 4x(√3x + 2) - √3(√3x + 2) = 0

⇒ (4x - √3) (√3x + 2) = 0

$\therefore \sf{x=\dfrac{\sqrt{3} }{4} \quad OR \quad x = \dfrac{-2}{\sqrt{3}}}$

$\rm{Let \: \alpha \: be \: \dfrac{\sqrt{3}}{4} \: and \: \beta \: be \: \dfrac{-2}{\sqrt{3}}}$

$\bf{\underline{Sum \: of \: zeroes:}}$

$\red{\sf{\alpha + \beta = \dfrac{-b}{a}}}$

$\to \: \sf{\dfrac{\sqrt{3}}{4} - \dfrac{2}{\sqrt{3}}= \dfrac{-5}{4 \sqrt{3}}}$

$\to \: \sf{\dfrac{3-8}{4\sqrt{3}}= \dfrac{-5}{4\sqrt{3}}}$

$\to \: \sf{\dfrac{-5}{4\sqrt{3}}=\dfrac{-5}{4\sqrt{3}}}$

LHS = RHS

$\\\bf{\underline{Product \: of \: zeros:}}$

$\sf{\pink{\alpha \beta = \dfrac{c}{a}}}$

$\to \: \sf{\dfrac{\sqrt{3}}{4}\times \dfrac{-2}{\sqrt{3}}=\dfrac{-2\sqrt{3}}{4\sqrt{3}}}$

$\to{\dfrac{-2}{4}=\dfrac{-2}{4}}$

LHS = RHS

★ Hence verified ★