Simplify: `(2x+p-c)^2-\ (2x-p+c)^2`

Simplify : p^(2)x - 2{px - 3x(x^(2) - bar(3a - x^(2)))}

Simplify : ((x^(a))/(x^(-b)))^(a^(2)-ab+b^(2))xx((x^(b))/(x^(-c)))^(b^(2)-bc+c^(2))xx((x^(c))/(x^(-a)))^(c^(2)-ca+a^(2))

Simplify the following (p+q) (p+2)

The straight line x+y=sqrt2P will touch the hyperbola 4x^2-9y^2=36 if (a) p^2=2 (b) p^2=5 (c) 5p^2=2 (d) p^2 = 2/ 5

In Fig. 4.220, D is the mid-point of side B C and A E_|_B C . If B C=a ,\ \ A C=b ,\ \ A B=c ,\ \ E D=x ,\ \ A D=p and A E=p and A E=h , prove that: (FIGURE) b^2=p^2+a x+(a^2)/4 (ii) c^2=p^2-a x+(a^2)/4 (iii) b^2+c^2=2\ p^2+(a^2)/2

Let a!=0 and p(x) be a polynomial of degree greater than 2. If p(x) leaves remainders a and - a when divided respectively, by x+a and x-a , the remainder when p(x) is divided by x^2-a^2 is (a) 2x (b) -2x (c) x (d) -x

Consider circles C_(1): x^(2) +y^(2) +2x - 2y +p = 0 C_(2): x^(2) +y^(2) - 2x +2y - p = 0 C_(3): x^(2) +y^(2) = p^(2) Statement-I: If the circle C_(3) intersects C_(1) orthogonally then C_(2) does not represent a circle Statement-II: If the circle C_(3) intersects C_(2) orthogonally then C_(2) and C_(3) have equal radii Then which of the following is true?

For p, q in R , the roots of (p^(2) + 2)x^(2) + 2x(p +q) - 2 = 0 are

If the quadratic equation a^(2)(b^(2) -c^(2)) x^(2) +b^(2) (c^(2) -a^(2)) x+c^(2) (a^(2) -b^(2)) =0 has real and equal roots, than a^(2) , b^(2) ,c^(2) are (1) A.P. (2) (G.P. (3) H.P. (4) A.G.P.

The base B C of a A B C is bisected at the point (p ,q) & the equation to the side A B&A C are p x+q y=1 & q x+p y=1 . The equation of the median through A is: (p-2q)x+(q-2p)y+1=0 (p+q)(x+y)-2=0 (2p q-1)(p x+q y-1)=(p^2+q^2-1)(q x+p y-1) none of these