Use suitable identities to find the following products:
(i) (x + 4) (x + 10)
(ii) (x + 8) (x – 10)
(iii) (3x + 4) (3x – 5)
(iv) \((y^ 2 + \frac{3 }{ 2}) (y^ 2 – \frac{3 }{ 2}) \)
(v) (3 – 2x) (3 + 2x)
(i) (x + 4)(x + 10) : By using the identity (x+a) (x+b) = x2 + (a+b)x + ab
(x+4) (x+10) = x2 + (4+10)x + 4 × 10 = x2 + 14x + 40
(ii) (x + 8)(x - 10) : By using the identity (x + a) (x + b) = x2 + (a + b)x + ab
(x+8) (x-10) = x2 + (x - 10)x + (8)(-10) = x2 - 2x - 80
(iii) (3x + 4)(3x - 5): 9(x + \(\frac{4 }{ 3}\))(x - \(\frac{5 }{ 3}\)) By using the identity (x + a) (x + b) = x2 + (a + b)x + ab
9(x + \(\frac{4 }{ 3}\))(x - \(\frac{5 }{ 3}\)) = 9[x2 + (\(\frac{4 }{ 3}\) - \(\frac{4 }{ 3}\))x + (\(\frac{4 }{ 3}\))(-\(\frac{5}{ 3}\))]
= 9[x2 - \(\frac{1 }{ 3}\)x - \(\frac{20 }{ 9}\)] = 9x2 - 3x - 20
(iv) (y2 + \(\frac{3 }{2}\))(y2 - \(\frac{3 }{ 2}\)) : By using the identity (x + y) (x - y) = x2 - y2
(y2 + \(\frac{3 }{ 2}\)) (y2 - \(\frac{3 }{ 2}\)) = (y2)2 - (\(\frac{3 }{ 2}\))2
= y4 - \(\frac{9 }{ 4}\)
(v) (3 - 2x)(3 + 2x) : By using the identity (x + y) (x - y) = x2 - y2
(3 - 2x) (3 + 2x) = (3)2 - (2x)2
= 9 - 4x2
If \( x = \left( 2 + \sqrt{3} \right)^3 + \left( 2 - \sqrt{3} \right)^{-3} \) and \( x^3 - 3x + k = 0 \), then the value of \( k \) is:
Two circles intersect at two points B and C. Through B, two line segments ABD and PBQ are drawn to intersect the circles at A, D and P, Q respectively (see Fig. 9.27). Prove that ∠ACP = ∠ QCD

ABCD is a trapezium in which AB || CD and AD = BC (see Fig. 8.14). Show that
(i) ∠A = ∠B
(ii) ∠C = ∠D
(iii) ∆ABC ≅ ∠∆BAD
(iv) diagonal AC = diagonal BD [Hint : Extend AB and draw a line through C parallel to DA intersecting AB produced at E.]

(i) The kind of person the doctor is (money, possessions)
(ii) The kind of person he wants to be (appearance, ambition)