Question

Find the zeroes of each of the following quadratic polynomials and verify the relationship

between the zeroes and their co efficient:

(i) f(x) =
2 − 2 − 8

(ii) g(s) = 4
2 − 4 + 1

(iii) h(t) =
2 − 15

(iv) p(x) =
2 + 2√2 + 6

(v) q(x) = √3
2 + 10 + 7√3

(vi) f(x) =
2 − (√3 + 1) + √3

(vii) g(x) = (
2 + 1) − (
2 + 1)

(viii) 6
2 − 3 − 7​

Answer 1

Answer:

(i) f(x) =  2 − 2 − 8 () =  2 − 2 − 8 =  2 − 4 + 2 − 8 = ( − 4) + 2( − 4) = ( + 2)( − 4) Zeroes of the polynomials are -2 and 4 Sum of the zeroes = −   -2 + 4 = −(−2) 1 2 = 2 Product of the zeroes =   2 = 24 = −8 1 − 8 = −8 ∴ Hence the relationship verified

(ii) 9(5) = 45 − 45 + 1 = 452 − 25 − 25 + 1 = 25(25 − 1) − 1(25 − 1) = (25 − 1)(25 − 1) Zeroes of the polynomials are 1 2 1 2 Sum of zeroes = −   2 1 2 + 1 2 = −(−4) 4 1 = 1 Product of the zeroes =   2 1 2 × 1 2 = 1 4 ⇒ 1 4 = 1 4 ∴ Hence the relationship verified.

(iii) h(t) =  2 − 15 = ( 2 ) − (√15) 2 = ( + √15)( − √15) zeroes of the polynomials are −√15 √15 sum of zeroes = 0 −√15 + √15 = 0 0 = 0Product of zeroes = −15 1 −√15 × √15 = −15 −15 = -15 ∴ Hence the relationship verified.

(iv) p(x) =  2 + 2√2 − 6 =  2 + 3√2 + √2 × 3√2 = ( + 3√2) − √2(2 + 3√2) = ( − √2)( + 3√2) Zeroes of the polynomial are 3√2 and -3√2 Sum of the zeroes = −3√2 1 √2 − 3√2 = −2√2 −2√2 = −2√2 ⇒ √2 × −3√2 = − 6 1 −6 = −6 ℎ ℎ 

(v) 2(x) = √3 2 + 10 + 7√3 = √3 2 + 7 + 3 + 7√3 = √3( + √3) + 7( + √3) = (√3 + 7)( + √3) Zeroes of the polynomials are −√3, −7 √3 Sum of zeroes = −10 √3 ⇒ −√3 − 7 √3 = −10 √3 ⇒ −10 √3 = −10 √3 Product of zeroes = 7√3 3 ⇒ √3−7 √30 = 7 ⇒ 7 = 7 Hence, relationship verified.

(vi) f(x) =  2 − (√3 + 1) + √3 =  2 − √3 − + √3 = x (x − √3) – 1 (x - √3) = (x – 1) (x − √3)  Zeroes of the polynomials are 1 and √3 Sum of zeroes = −{ }  2 = −[−√3−1] 1 1 + √3 = √3 + 1 Product of zeroes =   2 = √3 1 1 × √3 = √3 = √3 = √3 ∴ Hence, relationship verified

(vii) g(x) = [( 2 + 1) − ( 2 + 1)] 2 =  2 + −  2 −  =  2 − [( 2 + 1) − ] + 0 =  2 −  2 − + = ( − ) − 1( − ) = ( − )( − 1) Zeroes of the polynomials = 1   Sum of the zeroes = −[− 2−1]  ⇒ 1  + =  2+1  ⇒  2+1  =  2+1  Product of zeroes =   ⇒ 1  × =   ⇒  2+1  =  2+1  Product of zeroes =   ⇒ 1 = 1 Hence relationship verified

(viii) 6 2 − 3 − 7 = 6 2 − 7 − 3 = (3 + 11)(2 − 3) Zeroes of polynomials are +3 2 −1 3 Sum of zeroes = −1 3 + 3 2 = 7 6 = −(−7) 6 = −( )  2 Product of zeroes = −1 3 × 3 2 = −1 2 = −3 6 =   2 ∴ Hence, relationship verified.