How can we solve quadratic polynomials questions of class 10th math. It is very important question. Please guide me the best way for solving this question. Because its very important Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients. (i) x2–2x –8 (ii) 4s2–4s+1 (iii) 6×2–3–7x(iv) 4u2+8u(v) t2–15vi) 3×2–x–4

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# Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients. (i) x2–2x –8 (ii) 4s2–4s+1 (iii) 6×2–3–7x(iv) 4u2+8u(v) t2–15vi) 3×2–x–4 Q.1

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(i) x

^{2}–2x –8⇒x

^{2}– 4x+2x–8 = x(x–4)+2(x–4) = (x-4)(x+2)Therefore, zeroes of polynomial equation x

^{2}–2x–8 are (4, -2)Sum of zeroes = 4–2 = 2 = -(-2)/1 = -(Coefficient of x)/(Coefficient of x

^{2})Product of zeroes = 4×(-2) = -8 =-(8)/1 = (Constant term)/(Coefficient of x

^{2})(ii) 4s

^{2}–4s+1⇒4s

^{2}–2s–2s+1 = 2s(2s–1)–1(2s-1) = (2s–1)(2s–1)Therefore, zeroes of polynomial equation 4s

^{2}–4s+1 are (1/2, 1/2)Sum of zeroes = (½)+(1/2) = 1 = -4/4 = -(Coefficient of s)/(Coefficient of s

^{2})Product of zeros = (1/2)×(1/2) = 1/4 = (Constant term)/(Coefficient of s

^{2 })(iii) 6x

^{2}–3–7x⇒6x

^{2}–7x–3 = 6x^{2 }– 9x + 2x – 3 = 3x(2x – 3) +1(2x – 3) = (3x+1)(2x-3)Therefore, zeroes of polynomial equation 6x

^{2}–3–7x are (-1/3, 3/2)Sum of zeroes = -(1/3)+(3/2) = (7/6) = -(Coefficient of x)/(Coefficient of x

^{2})Product of zeroes = -(1/3)×(3/2) = -(3/6) = (Constant term) /(Coefficient of x

^{2 })(iv) 4u

^{2}+8u⇒ 4u(u+2)

Therefore, zeroes of polynomial equation 4u

^{2}+ 8u are (0, -2).Sum of zeroes = 0+(-2) = -2 = -(8/4) = = -(Coefficient of u)/(Coefficient of u

^{2})Product of zeroes = 0×-2 = 0 = 0/4 = (Constant term)/(Coefficient of u

^{2 })(v) t

^{2}–15⇒ t

^{2}= 15 or t = ±√15Therefore, zeroes of polynomial equation t

^{2}–15 are (√15, -√15)Sum of zeroes =√15+(-√15) = 0= -(0/1)= -(Coefficient of t) / (Coefficient of t

^{2})Product of zeroes = √15×(-√15) = -15 = -15/1 = (Constant term) / (Coefficient of t

^{2 })(vi) 3x

^{2}–x–4⇒ 3x

^{2}–4x+3x–4 = x(3x-4)+1(3x-4) = (3x – 4)(x + 1)Therefore, zeroes of polynomial equation3x

^{2}– x – 4 are (4/3, -1)Sum of zeroes = (4/3)+(-1) = (1/3)= -(-1/3) = -(Coefficient of x) / (Coefficient of x

^{2})Product of zeroes=(4/3)×(-1) = (-4/3) = (Constant term) /(Coefficient of x

^{2 })