If alpha and beta are the zeros of the polynomial

f(x) = 6𝑥2 − 𝑥 − 2 Sinceα andβ are the zeroes of the given polynomial ∴ Sum of zeroes [α + β] `=(-1)/6` Product of zeroes `(alphabeta)=(-1)/3` `=alpha/beta+beta/alpha=(alpha^2+beta^2)/(alphabeta)=((alpha+beta)^2-2alphabeta)/(alphabeta)` `=((1/6)^2-2xx((-1)/3))/(-1/3)` `=(1/6-2/3)/(-1/3)` `=((1+24)/36)/(-1/3)` `=(25/36)/(-1/3)` `=-25/12`

Concept : 1 Mark Application : 1 Mark Calculation : 2 Marks Since α and β are the zeros of the quadratic polynomial f ( x ) = k x 2 + 4 x + 4 ∴ α + β = − 4 k a n d α β = 4 k Now, α 2 + β 2 = 24 ⇒ ( α + β ) 2 − 2 α β = 24 ⇒ ( − 4 k ) 2 − 2 × 4 k = 24 ⇒ 16 k 2 − 8 k = 24 ⇒ 16 − 8 k = 24 k 2 ⇒ 3 k 2 + k − 2 = 0 ⇒ 3 k 2 + 3 k − 2 k − 2 = 0 ⇒ 3 k ( k + 1 ) − 2 ( k + 1 ) = 0 ⇒ ( k + 1 ) ( 3 k − 2 ) = 0 ⇒ k + 1 = 0 o r 3 k − 2 = 0 ⇒ k = − 1 o r k = 2 3 Hence, k = − 1 o r k = 2 3 .

1) x 2 − 2 x + 3 α + β = 2 α ⋅ β = 3 x 2 − ( ( α − 2 ) ( + ( β − 2 ) ) x + ( α − 2 ) ( β − 2 ) α − 2 + β − 2 = ( α + β − 4 ) ( α − 2 ) ( β − 2 ) = α β − 2 β − 2 α + 4 3 − 2 ( 2 ) + 4 = 3 x 2 + 2 x + 3 = 0. 2) α − 1 α + 1 + β − 1 β + 1 = ( α − 1 ) ( β + 1 ) + ( β − 1 ) ( α + 1 ) ( α + 1 ) β + 1 = 2 α β − 2 α β + α + β + 1 = 2 ⋅ 3 − 2 3 + 2 + 1 = 4 6 = 2 3 = ( α − 1 α + 1 ) ( β − 1 β + 1 ) = α β − β − α + 1 α β + α + β + 1 = 3 − ( 2 ) + 1 3 + 2 + 1 = 2 6 = 1 3 = x 2 − 2 3 x + 1 3 = 0 3 x 2 − 2 x + 1 = 0.

Given a quadratic polynomial #ax^2+bx+c# sum of its zeroes #alpha# and #beta# is #-b/a# and product of zeroes is #c/a#. Further if sum of zeroes is #s# and product of zeroes is #p#, then quadratic polynomial is #x^2-sx+p#. Hence for #3x^2+5x-2# we have #alpha+beta=-5/3# and #alphabeta=-2/3# We now desire quadratic polynomial whose zeroes are #2alpha# and #2beta# As sum of roots is #2alpha+2beta=2(alpha+beta)=2xx(-5)/3=-10/3# and product of roots is #2alphaxx2beta=4alphabeta=4xx(-2)/3=-8/3# and quadratic polynomial is #x^2+10/3x-8/3# as zeroes are not affected by multiplying each term of polynomial by a constant, we can say quadratic polynomial is #3x^2+10x-8#

Given alpha and beta are the zeroes of the polynomial kx^2+4x+4 I 'll denote alpha by x and beta by y for the sake of better to type and simplicity f(x)=ax 2+bx+c (general equation) equation f(x)=Kx 2+4x+4 sum of roots, x + y = -b/a (here b = -4 and a = 1) x+y=-4/k.......eq 1 product of roots , xy= c/a ( c = 4 and a = 1) xy= 4/k........eq 2 Squaring both sides of eq 1 (x+y)2= (-4/k) 2 x 2+y 2+2xy=16/k 2 24+2xy=16/k 2 [from x 2+y 2 = 24] Substituting the value of xy from eq2, 24+2*4/k=16/k 2 24+8/k=16/k 2 24k+8/k=16/k 2 24k+8=(16/k 2)*(k) 24k+8=16/k 24k 2+8k=16 or 24k 2+8k -16=0 3k 2+k-2=0 3k 2+3k-2k-2=0 3k(k+1)-2(k+1)=0 (3k-2)(k+1)=0 Therefore,k is either -1 or 2/3 Zeroes is otherwise known as roots of the equation. Q. (1) If alpha and beta are the zeroes of the polynomaial 2X(square)-5X+7 , Find a polynomial whose zeroes are 2 alpha + 3 beta and 3 alpha + 2 beta. (2) If alpha and beta are the zeroes of the polynimial 3X(square)-4X+1 , Find a polynomial whose zeroes are alpha(square) by (divided by ) beta and beta (square) by alpha. please help me....

If alpha and ß are zeroes of polynomial x²+6x+9 then alpha * ß(product of zeroes) =c/a =9/1 =9=>alpha * ß=9 ---(1) alpha * ß(sum of zeroes) =-b/a =-6/1 =-6 =>alpha + ß= -6 ----(2) Now zeroes of new quadratic equation are -alpha, -ß. sum of zeroes = -alpha+ -ß = -(alpha + ß) from (2) = --6 =6 ---(3) product of zeroes = -alpha * -ß = -alpha * -ß =alpha * ß from (1) = 9 ----(4) Quadratic equation if of the form x² -(sum of zeroes)x +product of zeroes =x² -(6)x + 9 =x² -6x + 9