Find the zeroes of the quadratic polynomial 4x2 - 8x - 6 and verify the relationship between zeroes and coefficients of the polynomial.

Let fx=4x2-8x-6For finding zeros , put fx=04x2-8x-6=02x2-4x-3=0x=--4±-42-42-322   using quadratic formulax=4±16+244x=4±404x=4±2104x=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

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4x2 – 8x – 6 = 0
2x2 – 4x – 3 = 0
Applying 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]/4
x =  [±√40 + 4]/4
x =  [±2√10 + 4]/4
x =  [±√10 + 2]/2
The relationship b/w roots & coefficients: {Where x1 & x2 are the roots}
A: x1 + x2 = -b/a
B: x1.x2 = c/a
Replacing:
A: L.H.S
[+√10 + 2]/2 + [-√10 + 2]/2
= [4/2]
= 2
R.H.S.
-(-8)/4
= 8/4
= 2
L.H.S = R.H.S
Hence verified (A)
B: L.H.S
{[+√10 + 2]/2}.{[-√10 + 2]/2}
= [+√10 + 2].[-√10 + 2]/4
= [4 - 10]/4
= -6/4
= -3/2
R.H.S
-6/2
= -3/2
L.H.S = R.H.S
Hence verified (B)
 
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