Find a quadratic polynomial each with the given numbers as the sum and product of its
zeroes respectively. (a) 4,0 (b) 2/3, 1/3
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Step-by-step explanation:
respectively.(i) 14 , - 1 (ii) √(2), 13 (iii) 0, √(5) (iv) 1, 1 (v) -14, 14 (vi) 4, 1
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(i) 14 , - 1Using the quadratic equation formula, x^2 - (Sum of root)x + (Product of root) = 0 Substitute the value in the formula, we get x^2 - 14x - 1 = 0 4x^2 - x - 4 = 0 (ii) √(2), 13 Using the quadratic equation formula, x^2 - (Sum of root)x + (Product of root) = 0 Substitute the value in the formula, we get x^2 - √(2)x + 13 = 0 Multiply by 3 to remove denominator, 3x^2 - 3√(2)x + 1 = 0 (iii) 0, √(5) Using the quadratic equation formula, x^2 - (Sum of root)x + (Product of root) = 0 Substitute the value in the formula, we get x^2 - 0x + √(5) = 0 x^2 + √(5) = 0 (iv) 1, 1Using the quadratic equation formula, x^2 - (Sum of root)x + (Product of root) = 0 Substitute the value in the formula, we get x^2 - 1x + 1 = 0 x^2 - x + 1 = 0 (v) -14, 14 Using the quadratic equation formula, x^2 - (Sum of root)x + (Product of root) = 0 Substitute the value in the formula, we get x^2 - -14x + 14 = 0 Multiply by 4 4x^2 + x + 1 = 0 (vi) 4, 1Using the quadratic equation formula, x^2 - (Sum of root)x +
Solution: (i)1/4, -1 Now formula of quadratic equation is x²-(Sum of root)x + (Product of root) = 0 Plug the value in formula we get x² –(1/4)x -1 = 0 Multiply by 4 to remove denominator we get 4x² - x -4 = 0 (ii) √2 , 1/3 Now formula of quadratic equation is x²-(Sum of root)x + (Product of root) = 0 Plug the value in formula we get x² –(√2)x + 1/3 = 0 Multiply by 3 to remove denominator we get 3x² - 3√2 x + 1 = 0 (iii) 0, √5 Now formula of quadratic equation is x²-(Sum of root)x + (Product of root) = 0 Plug the value in formula we get x² –(0)x + √5 = 0 simplify it we get x² + √5 = 0 (iv) 1,1 Now formula of quadratic equation is x²-(Sum of root)x + (Product of root) = 0 Plug the value in formula we get x² –(1)x + 1 = 0 simplify it we get x² - x + 1 = 0 (v) -1/4 ,1/4 Now formula of quadratic equation is x²-(Sum of root)x + (Product of root) = 0 Plug the value in formula we get x² –(-1/4)x + 1/4 = 0 multiply by 4 we get 4x² + x + 1 = 0 (vi) 4,1 Now formula of quadratic equation is…
Step-by-step explanation:(vi) 4,1sum of zeros =4product of zeros=1R.P=K{x2-(sum)x+(product)} =K{x2-(4)x+(1)} =K{x2-4x+1}hence,the quadratic polynomial is x2-4x+1
We have, - b/a = 1/4 b/a = - 1/4 c/a = - 1 Since, ax^2 + bx + c = 0 be a polynomial. x^2 + ba x + ca = 0 x^2 - 14 x - 1 = 0 4x^2 - x - 4 = 0 Hence, this is the answer.
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Find a quadratic polynomial each with the given number as the sum and product of its zeroes respectively. : -3,2
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Formula for quadratic equation is,x2 – (sum of roots) x + Product of roots = 0Given, sum of roots = and product of roots = ∴ Quadratic equation is,⇒ x2 – x – = 0⇒ = 0Hence, Quadratic equation is = 0.
Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively. (i)1/4 , -1 (ii) √2 , 1/3
[i] 1/4(4x² + x + 1)[ii] 1/3(3x² - 3√2x + 1)Justification:Let α, β be the zeros of polynomial.[i] We have, α + β = - 1/4 & αβ = 1/4Thus, polynomial isp(x) = x² - (α + β)x + αβx² - (-1/4)x + 1/4x² + 1/4x + 1/41/4(4x² + x + 1)[ii] We have, α + β = √2 & αβ = 1/3Thus, polynomial isp(x) = x² - (α + β)x + αβx² - √2x + 1/31/3(3x² - 3√2x + 1)
assume a=3 as one root and b=-2/5 as the other. Sum of roots=3+(-2/5)=13/5 Product of roots=3*-2/5= -6/5 Therefore, we form the equation using- x^2 - (sum of roots)*x + (product of roots) = 0 x^2 - (13/5)*x + (-6/5) = 0 x^2 - 13x/5 - 6/5 = 0 if we take LCM and simplify, we finally get- 5x^2– 13x - 6 = 0. Thank you.