Question

find the zeros of the quadratic polynomial 4 root 3 x2 +5x-2 root3 and verify the relationship between the zeros and the coefficients

Answer 1

Solution :

We have quadratic polynomial p(x) = 4√3x² + 5x -2√3 & zero of the polynomial p(x) = 0

$\underline{\underline{\tt{Using\:\:by\:\:Factorization\:\:method\::}}}$

$\longrightarrow\sf{4\sqrt{3} x^{2} + 5x -2\sqrt{3} =0}\\\\\longrightarrow\sf{4\sqrt{3} x^{2} +8x-3x -2\sqrt{3} =0}\\\\\longrightarrow\sf{4x(\sqrt{3} x+2 ) -\sqrt{3} (\sqrt{3} x + 2) = 0}\\\\\longrightarrow\sf{(\sqrt{3} x + 2) (4x -\sqrt{3}) = 0}\\ \\\longrightarrow\sf{\sqrt{3} x + 2 =0\:\:\:Or\:\:\:4x-\sqrt{3} =0}\\\\\longrightarrow\sf{\sqrt{3} x =-2\:\:\:Or\:\:\:4x=\sqrt{3} }\\\\\longrightarrow\bf{ x =-2/\sqrt{3} \:\:\:Or\:\:\:x=\sqrt{3} /4}$

∴ α = -2/√3 & β = √3/4 are the zeroes of the given polynomial .

As we know that given quadratic polynomial compared with ax² + bx + c;

• a = 4√3
• b = 5
• c = -2√3

Now;

$\underline{\mathcal{SUM\:OF\:THE\:ZEROES\::}}$

$\mapsto\tt{\alpha + \beta = \dfrac{-b}{a} =\bigg\lgroup\dfrac{Coefficient\:of\:x}{Coefficient\:of\:x^{2}} \bigg\rgroup}\\\\\\\mapsto\tt{-\dfrac{2}{\sqrt{3} } +\dfrac{\sqrt{3} }{4} =\dfrac{-5}{4\sqrt{3} }} \\\\\\\mapsto\tt{\dfrac{-8+3}{4\sqrt{3} } =\dfrac{-5}{4\sqrt{3}} } \\\\\\\mapsto\bf{\dfrac{-5}{4\sqrt{3} } =\dfrac{-5}{4\sqrt{3} }}$

$\underline{\mathcal{PRODUCT\:OF\:THE\:ZEROES\::}}$

$\mapsto\tt{\alpha \times \beta = \dfrac{c}{a} =\bigg\lgroup\dfrac{Constant\:term}{Coefficient\:of\:x^{2}} \bigg\rgroup}\\\\\\\mapsto\tt{-\dfrac{2}{\sqrt{3} } \times \dfrac{\sqrt{3} }{4} =\dfrac{-2\sqrt{3} }{4\sqrt{3} }} \\\\\\\mapsto\bf{\dfrac{-2\sqrt{3} }{4\sqrt{3} } =\dfrac{-2\sqrt{3} }{4\sqrt{3} }}$

Thus;

The relationship between zeroes & coefficient are verified .