If the zeroes of the polynomial (x^2) +px +q are double in value to the zeroes of the polynomial 2(x^2 )-5x -3, then find the values of p and q.

 If the zeroes of the polynomial (x^2) +px +q are double in value to the zeroes of the
polynomial 2(x^2 )-5x -3, then find the values of p and q.

 

 given expression 2(x^2 )-5x -3

a=2

b = (-5)

c=(-3)

 

using formulae if alpha and beta are the ZEROES of the given polynomial

alpha + beta = (-b)/a = -(-5)/2=5/2

alpha*beta =c/a =(-3)/2

given that the zerroes of

(x^2) +px +q are double in value to the zeroes of the
polynomial 2(x^2 )-5x -3

 

assume the new zerroes are 2alpha and 2beta

sum of new zeroes is 2alpha +2beta = 2(alpha+beta) = 2(5/2)=5

product of the new zeroes are  2alpha *2beta = 4*(alpha*beta) =4*(-3)/2 = (-6)

new polynomial is (x^2) -(sum)x +product

(x^2) -5x -6

compare

(x^2) +px +q

 

to get p =(-5) and q=(-6) 

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Find an acute angle θ when (cosθ − sin θ)/(cosθ+sin θ) = (1−√3)/(1+√3)

 Find an acute angle θ when (cosθ − sin θ)/(cosθ+sin θ) = (1−√3)/(1+√3)

 

given  (cosθ − sin θ)/(cosθ+sin θ) = (1−√3)/(1+√3)

 divide each term with cosθ on the LHS and use sinθ/cosθ =tanθ

we get

(1-tan θ)/(1+tanθ) = (1−√3)/(1+√3)

 rearrange and simplify

we get tan θ = √3

 θ =60 degrees

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If sin(A+B) =1 and cos(A-B)= √3/2, 0°< A+B ≤ 90° and A> B, then find the measures of angles A and B.

 If sin(A+B) =1 and cos(A-B)= √3/2, 0°< A+B ≤ 90° and A> B, then find the
measures of angles A and B.

given sin(A+B) =1=sin(90 degrees)

so

 A+B =90degreees-------------(1)

given cos(A-B)= √3/2=cos(30 degrees)

 

A-B = 30 degrees-----------------(2)

 

now solve the two equations

 

adding the two equations gives 2A =120 or A = 60 degrees

 

substitute in the first equation to get B= 30 degrees 

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In the given figure, arcs have been drawn of radius 7cm each with vertices A, B, C and D of quadrilateral ABCD as centres. Find the area of the shaded region.

In the given figure, arcs have been drawn of radius 7cm each with vertices A, B, C
and D of quadrilateral ABCD as centres. Find the area of the shaded region consisting of the four sectors at the four vertices

 

in a quadrilateral sum of the interior angles is 360 degrees

so A+B+C+D =360 DEGREES

radius =7 cm

 area of the sector = (theta /360)*pi* (r^2)

 

so sum of the areas of the sectors = ( (A+B+C+D) /360 )*pi *(r^2)

=(360/360)*(22/7)*7*7=84 square centimetre
 

 

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