Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

1. Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.

1. Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.

Grade:12th pass

1 Answers

Pawan Prajapati
askIITians Faculty 8740 Points
4 months ago
Solutions: (i) x2–2x –8 ⇒x2– 4x+2x–8 = x(x–4)+2(x–4) = (x-4)(x+2) Therefore, zeroes of polynomial equation x2–2x–8 are (4, -2) Sum of zeroes = 4–2 = 2 = -(-2)/1 = -(Coefficient of x)/(Coefficient of x2) Product of zeroes = 4×(-2) = -8 =-(8)/1 = (Constant term)/(Coefficient of x2) (ii) 4s2–4s+1 ⇒4s2–2s–2s+1 = 2s(2s–1)–1(2s-1) = (2s–1)(2s–1) Therefore, zeroes of polynomial equation 4s2–4s+1 are (1/2, 1/2) Sum of zeroes = (½)+(1/2) = 1 = -4/4 = -(Coefficient of s)/(Coefficient of s2) Product of zeros = (1/2)×(1/2) = 1/4 = (Constant term)/(Coefficient of s2 ) (iii) 6x2–3–7x ⇒6x2–7x–3 = 6x2 – 9x + 2x – 3 = 3x(2x – 3) +1(2x – 3) = (3x+1)(2x-3) Therefore, zeroes of polynomial equation 6x2–3–7x are (-1/3, 3/2) Sum of zeroes = -(1/3)+(3/2) = (7/6) = -(Coefficient of x)/(Coefficient of x2) Product of zeroes = -(1/3)×(3/2) = -(3/6) = (Constant term) /(Coefficient of x2 ) (iv) 4u2+8u ⇒ 4u(u+2) Therefore, zeroes of polynomial equation 4u2 + 8u are (0, -2). Sum of zeroes = 0+(-2) = -2 = -(8/4) = = -(Coefficient of u)/(Coefficient of u2) Product of zeroes = 0×-2 = 0 = 0/4 = (Constant term)/(Coefficient of u2 ) (v) t2–15 ⇒ t2 = 15 or t = ±√15 Therefore, zeroes of polynomial equation t2 –15 are (√15, -√15) Sum of zeroes =√15+(-√15) = 0= -(0/1)= -(Coefficient of t) / (Coefficient of t2) Product of zeroes = √15×(-√15) = -15 = -15/1 = (Constant term) / (Coefficient of t2 ) (vi) 3x2–x–4 ⇒ 3x2–4x+3x–4 = x(3x-4)+1(3x-4) = (3x – 4)(x + 1) Therefore, zeroes of polynomial equation3x2 – x – 4 are (4/3, -1) Sum of zeroes = (4/3)+(-1) = (1/3)= -(-1/3) = -(Coefficient of x) / (Coefficient of x2) Product of zeroes=(4/3)×(-1) = (-4/3) = (Constant term) /(Coefficient of x2 )

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

ASK QUESTION

Get your questions answered by the expert for free