Question 1. Find the zeros of each of the following quadratic polynomials and verify the relationship between the zeros and their coefficients:(i) f(x) = x2 – 2x – 8Solution: Given that, f(x) = x2 – 2x – 8 To find the zeros of the equation, put f(x) = 0 = x2 – 2x – 8 = 0 = x2 – 4x + 2x – 8 = 0 = x(x – 4) + 2(x – 4) = 0 = (x – 4)(x + 2) = 0 x = 4 and x = -2 Hence, the zeros of the quadratic equation are 4 and -2. Now, Verification As we know that, Sum of zeros = – coefficient of x / coefficient of x^2 4 + (-2)= – (-2) / 1 2 = 2 Product of roots = constant / coefficient of x^2 4 x (-2) = (-8) / 1 -8 = -8 Hence the relationship between zeros and their coefficients are verified.
(ii) g(s) = 4s2 – 4s + 1Solution: Given that, g(s) = 4s2 – 4s + 1 To find the zeros of the equation, put g(s) = 0 = 4s2 – 4s + 1 = 0 = 4s2 – 2s – 2s + 1= 0 = 2s(2s – 1) – (2s – 1) = 0 = (2s – 1)(2s – 1) = 0 s = 1/2 and s = 1/2 Hence, the zeros of the quadratic equation are 1/2 and 1/2. Now, Verification As we know that, Sum of zeros = – coefficient of s / coefficient of s2 1/2 + 1/2 = – (-4) / 4 1 = 1 Product of roots = constant / coefficient of s2 1/2 x 1/2 = 1/4 1/4 = 1/4 Hence the relationship between zeros and their coefficients are verified.
(iii) h(t)=t2 – 15Solution: Given that, h(t) = t2 – 15 = t2 +(0)t – 15 To find the zeros of the equation, put h(t) = 0 = t2 – 15 = 0 = (t + √15)(t – √15)= 0 t = √15 and t = -√15 Hence, the zeros of the quadratic equation are √15 and -√15. Now, Verification As we know that, Sum of zeros = – coefficient of t / coefficient of t2 √15 + (-√15) = – (0) / 1 0 = 0 Product of roots = constant / coefficient of t2 √15 x (-√15) = -15/1 -15 = -15 Hence the relationship between zeros and their coefficients are verified.
(iv) f(x) = 6x2 – 3 – 7xSolution: Given that, f(x) = 6x2 – 3 – 7x To find the zeros of the equation, we put f(x) = 0 = 6x2 – 3 – 7x = 0 = 6x2 – 9x + 2x – 3 = 0 = 3x(2x – 3) + 1(2x – 3) = 0 = (2x – 3)(3x + 1) = 0 x = 3/2 and x = -1/3 Hence, the zeros of the quadratic equation are 3/2 and -1/3. Now, Verification As we know that, Sum of zeros = – coefficient of x / coefficient of x2 3/2 + (-1/3) = – (-7) / 6 7/6 = 7/6 Product of roots = constant / coefficient of x2 3/2 x (-1/3) = (-3) / 6 -1/2 = -1/2 Hence the relationship between zeros and their coefficients are verified.
(v) p(x) = x2 + 2√2x – 6Solution: Given that, p(x) = x2 + 2√2x – 6 To find the zeros of the equation, put p(x) = 0 = x2 + 2√2x – 6 = 0 = x2 + 3√2x – √2x – 6 = 0 = x(x + 3√2) – √2 (x + 3√2) = 0 = (x – √2)(x + 3√2) = 0 x = √2 and x = -3√2 Hence, the zeros of the quadratic equation are √2 and -3√2. Now, Verification As we know that, Sum of zeros = – coefficient of x / coefficient of x2 √2 + (-3√2) = – (2√2) / 1 -2√2 = -2√2 Product of roots = constant / coefficient of x2 √2 x (-3√2) = (-6) / 2√2 -3 x 2 = -6/1 -6 = -6 Hence the relationship between zeros and their coefficients are verified.
(vi) q(x)=√3x2 + 10x + 7√3Solution: Given that, q(x) = √3x2 + 10x + 7√3 To find the zeros of the equation, put q(x) = 0 = √3x2 + 10x + 7√3 = 0 = √3x2 + 3x +7x + 7√3x = 0 = √3x(x + √3) + 7 (x + √3) = 0 = (x + √3)(√3x + 7) = 0 x = -√3 and x = -7/√3 Hence, the zeros of the quadratic equation are -√3 and -7/√3. Now, Verification As we know that, Sum of zeros = – coefficient of x / coefficient of x2 -√3 + (-7/√3) = – (10) /√3 (-3-7)/ √3 = -10/√3 -10/ √3 = -10/√3 Product of roots = constant / coefficient of x2 (-√3) x (-7/√3) = (7√3) / √3 7 = 7 Hence the relationship between zeros and their coefficients are verified.
(vii) f(x) = x2 – (√3 + 1)x + √3Solution: Given that, f(x) = x2 – (√3 + 1)x + √3 To find the zeros of the equation, put f(x) = 0 = x2 – (√3 + 1)x + √3 = 0 = x2 – √3x – x + √3 = 0 = x(x – √3) – 1 (x – √3) = 0 = (x – √3)(x – 1) = 0 x = √3 and x = 1 Hence, the zeros of the quadratic equation are √3 and 1. Now, Verification Sum of zeros = – coefficient of x / coefficient of x2 √3 + 1 = – (-(√3 +1)) / 1 √3 + 1 = √3 +1 Product of roots = constant / coefficient of x2 1 x √3 = √3 / 1 √3 = √3 Hence the relationship between zeros and their coefficients are verified.
(viii) g(x) = a(x2+1)–x(a2+1)Solution: Given that, g(x) = a(x2+1)–x(a2+1) To find the zeros of the equation put g(x) = 0 = a(x2+1)–x(a2+1) = 0 = ax2 + a − a2x – x = 0 = ax2 − a2x – x + a = 0 = ax(x − a) − 1(x – a) = 0 = (x – a)(ax – 1) = 0 x = a and x = 1/a Hence, the zeros of the quadratic equation are a and 1/a. Now, Verification : As we know that, Sum of zeros = – coefficient of x / coefficient of x2 a + 1/a = – (-(a2 + 1)) / a (a^2 + 1)/a = (a2 + 1)/a Product of roots = constant / coefficient of x2 a x 1/a = a / a 1 = 1 Hence the relationship between zeros and their coefficients are verified.
(ix) h(s) = 2s2 – (1 + 2√2)s + √2Solution: Given that, h(s) = 2s2 – (1 + 2√2)s + √2 To find the zeros of the equation put h(s) = 0 = 2s2 – (1 + 2√2)s + √2 = 0 = 2s2 – 2√2s – s + √2 = 0 = 2s(s – √2) -1(s – √2) = 0 = (2s – 1)(s – √2) = 0 x = √2 and x = 1/2 Hence, the zeros of the quadratic equation are √3 and 1. Now, Verification As we know that, Sum of zeros = – coefficient of s / coefficient of s2 √2 + 1/2 = – (-(1 + 2√2)) / 2 (2√2 + 1)/2 = (2√2 +1)/2 Product of roots = constant / coefficient of s2 1/2 x √2 = √2 / 2 √2 / 2 = √2 / 2 Hence the relationship between zeros and their coefficients are verified.
(x) f(v) = v2 + 4√3v – 15Solution: Given that, f(v) = v2 + 4√3v – 15 To find the zeros of the equation put f(v) = 0 = v2 + 4√3v – 15 = 0 = v2 + 5√3v – √3v – 15 = 0 = v(v + 5√3) – √3 (v + 5√3) = 0 = (v – √3)(v + 5√3) = 0 v = √3 and v = -5√3 Hence, the zeros of the quadratic equation are √3 and -5√3. Now, for verification Sum of zeros = – coefficient of v / coefficient of v2 √3 + (-5√3) = – (4√3) / 1 -4√3 = -4√3 Product of roots = constant / coefficient of v2 √3 x (-5√3) = (-15) / 1 -5 x 3 = -15 -15 = -15 Hence the relationship between zeros and their coefficients are verified.
(xi) p(y) = y2 + (3√5/2)y – 5Solution: Given that, p(y) = y2 + (3√5/2)y – 5 To find the zeros of the equation put f(v) = 0 = y2 + (3√5/2)y – 5 = 0 = y2 – √5/2 y + 2√5y – 5 = 0 = y(y – √5/2) + 2√5 (y – √5/2) = 0 = (y + 2√5)(y – √5/2) = 0 This gives us 2 zeros, y = √5/2 and y = -2√5 Hence, the zeros of the quadratic equation are √5/2 and -2√5. Now, Verification As we know that, Sum of zeros = – coefficient of y / coefficient of y2 √5/2 + (-2√5) = – (3√5/2) / 1 -3√5/2 = -3√5/2 Product of roots = constant / coefficient of y2 √5/2 x (-2√5) = (-5) / 1 – (√5)2 = -5 -5 = -5 Hence the relationship between zeros and their coefficients are verified.
(xii) q(y) = 7y2 – (11/3)y – 2/3Solution: Given that, q(y) = 7y2 – (11/3)y – 2/3 To find the zeros of the equation put q(y) = 0 = 7y2 – (11/3)y – 2/3 = 0 = (21y2 – 11y -2)/3 = 0 = 21y2 – 11y – 2 = 0 = 21y2 – 14y + 3y – 2 = 0 = 7y(3y – 2) – 1(3y + 2) = 0 = (3y – 2)(7y + 1) = 0 y = 2/3 and y = -1/7 Hence, the zeros of the quadratic equation are 2/3 and -1/7. Now, Verification As we know that, Sum of zeros = – coefficient of y / coefficient of y2 2/3 + (-1/7) = – (-11/3) / 7 -11/21 = -11/21 Product of roots = constant / coefficient of y2 2/3 x (-1/7) = (-2/3) / 7 – 2/21 = -2/21 Hence the relationship between zeros and their coefficients are verified.
Question 2. For each of the following, find a quadratic polynomial whose sum and product respectively of the zeros are as given. Also, find the zeros of these polynomials by factorization.(i) -8/3, 4/3Solution: As we know that the quadratic polynomial formed for the given sum and product of zeros is given by : f(x) = x2 + -(sum of zeros) x + (product of roots)
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