Question

Find the zeros of the quadratic polynomial √3x^2+5/4x-√3/2 and verify the relationship between the zeros and its coefficients

Answer 1

Answer :

Given quadratic equation is   $\sqrt{3}x^2+\frac{5}{4}x-\frac{\sqrt{3} }{2}$

$\bold{\sqrt{3}x^2+\frac{5}{4}x-\frac{\sqrt{3} }{2} } \\\\\implies \bold{\frac{4\sqrt{3}x^2+5x-2\sqrt{3} }{4} }\\\\\implies \bold{\frac{4\sqrt{3}x^2+8x-3x-2\sqrt{3} }{4} }\\\\\implies \bold{\frac{4x(\sqrt{3}x+2)-\sqrt{3}(\sqrt{3}x+2 ) )}{4} }\\\\\implies \bold{\frac{(\sqrt{3}x+2)(4x-\sqrt{3} )}{4} =0}\\\\\implies \bold{\;x=-\frac{2}{\sqrt{3} } \;,\;x=\frac{\sqrt{3}}{4} }$

Compare given equation $\bold{\sqrt{3}x^2+\frac{5}{4}x-\frac{\sqrt{3} }{2} }$ with ax²+bx+c , we get ,

• a = √3 , b = 5/4 , c = -√3/2

Verification :

$\bold{Sum\;of\;roots\;,\alpha +\beta =\frac{-b}{a} }\\\\\bold{\implies \frac{-2}{\sqrt{3} } +\frac{\sqrt{3} }{4} =\frac{\frac{-5}{4} }{\sqrt{3}} }\\\\\bold{\implies \frac{-8+3}{4\sqrt{3} } =\frac{-5}{4\sqrt{3} } }\\\\\bold{\implies \frac{-5}{4\sqrt{3} } =\frac{-5}{4\sqrt{3} } }$

$\bold{Product\;of\;roots\;,\;\alpha \beta =\frac{c}{a} }\\\\\implies \bold{\frac{-2}{\sqrt{3} } .\frac{\sqrt{3} }{4}=\frac{\frac{\sqrt{3} }{2} }{\sqrt{3} } }\\\\\implies \bold{-\frac{1}{2} =-\frac{1}{2} }$

Hence Verified .