Find the zeroes of the quadratic polynomial 4x2 - 8x - 6 and verify the relationship between zeroes and coefficients of the polynomial. Share with your friends Share 1 Neha Sethi answered this Let fx=4x2-8x-6For finding zeros , put fx=0⇒4x2-8x-6=0⇒2x2-4x-3=0x=--4±-42-42-322 using quadratic formula⇒x=4±16+244⇒x=4±404⇒x=4±2104⇒x=2±102Verification:Sum of roots=2+102+2-102=42=2and -coeff. of xcoeff of x2=--84=2So, Sum of roots=-coeff. of xcoeff of x2Product of roots=2+102×2-102=4-104=-64=-32and constant termcoeff. of x2=-64=-32So, Product of roots=constant termcoeff. of x2 3 View Full Answer Priyamtejpal answered this 4x2 – 8x – 6 = 02x2 – 4x – 3 = 0Applying the quadratic formula: {x = [±√(b2 – 4ac) - b]/2a where a, b & c are respectively the coefficients of x2 , x1, & x0}:x = [±√((-4)2 – 4(2)(-3)) – (-4)]/2(2)x = [±√(16 + 24) + 4]/4x = [±√40 + 4]/4x = [±2√10 + 4]/4x = [±√10 + 2]/2The relationship b/w roots & coefficients: {Where x1 & x2 are the roots}A: x1 + x2 = -b/aB: x1.x2 = c/aReplacing:A: L.H.S[+√10 + 2]/2 + [-√10 + 2]/2= [4/2]= 2R.H.S.-(-8)/4= 8/4= 2L.H.S = R.H.SHence verified (A)B: L.H.S{[+√10 + 2]/2}.{[-√10 + 2]/2}= [+√10 + 2].[-√10 + 2]/4= [4 - 10]/4= -6/4= -3/2R.H.S-6/2= -3/2L.H.S = R.H.SHence verified (B) -2