polynomials

Q)  Find the zeroes of the polynomial f(t) = t2 + 4 √3 t – 15 and verify the relationship between the zeroes and the coefficients of the polynomial. Ans: In the given polynomial equation, to find zeroes, we will start with f(x) = 0 Therefore, t2 + 4√3 t – 15 = 0 Step

Q)  If 𝛼, β are zeroes of quadratic polynomial f(x) = 6 x2 + 11 x – 10, find the value of Ans: In the given polynomial equation,  to find zeroes, we will start with f(x) = 0 Therefore, 6 x2 + 11 x – 10 = 0 Step 1: Given that the roots of

Q. Solve for X : Ans: Given equation is: ∴ ∴ 4 (2 X + 3) – 5 X = 3 X (2X + 3) ∴ 8 X + 12 – 5 X = 6 X 2  + 9 X ∴ 12  = 6 X 2  + 6 X ∴ 6 X 2  + 6 X

Q. The sum of two numbers is 18 and the sum of their reciprocals is 1/4 . Find the numbers. Ans: Let the numbers be X and Y Step 1: By 1st condition: X + Y = 18 or X = 18 – Y ….. (i) Step 2: By 2nd condition: By simplifying this equation, we

Q)  The graph of the polynomial – ax²+bx+c is shown below. Write the signs of ‘a’ and ‘b² – 4ac’. Ans: Clearly, f(x) = ax² + bx + c represent a parabola opening upwards. Therefore, a>0 Since, the parabola cuts the x-axis at two points, this means that the polynomial will have two real solutions. Hence,

Q)  Find the value of m for which the quadratic equation (m + 1) y2 – 6(m+1) y + 3(m+9) = 0, m ≠ -1 has equal roots. Hence find the roots of the equation. Ans: Given quadratic equation is: (m + 1) y2 – 6(m+1) y + 3(m+9) = 0 If we compare to ax2

Q)  If 𝛼, β are zeroes of quadratic polynomial x2 – 2x + 3, find the polynomial whose roots are:1. 𝛼 + 2, 𝛽 + 22. Ans: Given polynomial equation x2 – 2x + 3 = 0 Comparing with standard polynomial, ax2 + b x + c = 0, we get, a =  1, b =