Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.
(i) $x^{2} - 2 x - 8$ (ii) $4 s^{2} - 4 s + 1$ (iii) $6 x^{2} - 3 - 7 x$
(iv) $4 u^{2} + 8 u$ (v) $t^{2} - 15$ (vi) $3 x^{2} - x - 4$

Question

Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.
(i)

x^{2} - 2 x - 8

(ii)

4 s^{2} - 4 s + 1

(iii)

6 x^{2} - 3 - 7 x

(iv)

4 u^{2} + 8 u

(v)

t^{2} - 15

(vi)

3 x^{2} - x - 4

Filo · Accepted Answer

(i) x2−2x−8Factorize the equation, we get (x+2)(x−4)So, the value of x2−2x−8 is zero when x+2=0,x−4=0, i.e., when x=−2 or x=4.Therefore, the zeros of x2−2x−8 are -2 and 4.Now,⇒Sum of zeroes =−2+4=2=−12​=−Coefficient  of  x2Coefficient  of  x​⇒Product of zeros =(−2)×(4)=−8 = 1−8​=Coefficient  of  x2Constant  term​(ii) 4s2−4s+1Factorize the equation, we get(2s−1)(2s−1)So, the value of 4s2−4s+1 is zero when 2s−1=0,2s−1=0, i.e., when s=21​ or s=21​.Therefore, the zeros of 4s2−4s+1 are 21​ and 21​.Now,⇒Sum of zeroes = 21​+21​=1=−4−4​=−Coefficient  of  s2Coefficient  of  s​⇒Product of zeros = 21​×21​=41​=41​=Coefficient  of  s2Constant  term​(iii) 6x2−3−7x Factorize the equation, we get (3x+1)(2x−3)So, the value of 6x2−3−7x is zero when 3x+1=0,2x−3=0, i.e., when x=−31​ or x=23​.Therefore, the zeros of 6x2−3−7x are −31​ and 23​.Now,⇒Sum of zeroes = −31​+23​=67​=−6−7​=−Coefficient  of  x2Coefficient  of  x​⇒Product of zeros = −31​×23​=−21​=6−3​=2−1​=Coefficient  of  x2Constant  term​(iv) 4u2+8u  Factorize the equation, we get 4u(u+2)So, the value of 4u2+8u is zero when 4u=0,u+2=0, i.e., when u=0 or u=−2.Therefore, the zeros of 4u2+8u are 0 and −2.Now,⇒Sum of zeroes = 0−2=−2=−48​= −2=−Coefficient  of  u2Coefficient  of  u​⇒Product of zeros = −0x−2=0=40​=0=Coefficient  of  u2Constant  term​(v) t2−15Factorize the equation, we get t=±15​So, the value of t2−15 is zero when t+15​=0,t−15​=0, i.e., when t=15​ or t=−15​.Therefore, the zeros of t2−15 are ±15​.Now,⇒Sum of zeroes = 15​−15​=0= −10​=0= −Coefficient  of  t2Coefficient  of  t​⇒Product of zeros = 15​×−15​=−15=1−15​=Coefficient  of  t2Constant  term​(vi) 3x2−x−4Factorize the equation, we get(x+1)(3x−4)So, the value of 3x2−x−4 is zero when x + 1 = 0, 3x - 4 = 0, i.e., when x = -1 or x = 34​.Therefore, the zeros of 3x2−x−4 are -1 and 34​.Now,⇒Sum of zeroes = −1+34​=31​= −3−1​= −Coefficient  of  x2Coefficient  of  x​⇒Product of zeros = −1×34​=−34​=3−4​=Coefficient  of  x2Constant  term​

Question Text	Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients. (i) $x^{2} - 2 x - 8$ (ii) $4 s^{2} - 4 s + 1$ (iii) $6 x^{2} - 3 - 7 x$ (iv) $4 u^{2} + 8 u$ (v) $t^{2} - 15$ (vi) $3 x^{2} - x - 4$
Answer Type	Text solution:1
Upvotes	150

Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.
(i) $x^{2} - 2 x - 8$ (ii) $4 s^{2} - 4 s + 1$ (iii) $6 x^{2} - 3 - 7 x$
(iv) $4 u^{2} + 8 u$ (v) $t^{2} - 15$ (vi) $3 x^{2} - x - 4$

Text solutionVerified

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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) 6x2−3−7x (iv) 4u2+8u (v) t2−15 (vi) 3x2−x−4

Text solutionVerified

Practice questions from similar books

Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.
(i) $x^{2} - 2 x - 8$ (ii) $4 s^{2} - 4 s + 1$ (iii) $6 x^{2} - 3 - 7 x$
(iv) $4 u^{2} + 8 u$ (v) $t^{2} - 15$ (vi) $3 x^{2} - x - 4$