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:
A driver of a car travelling at \(52\) \(km \;h^{–1}\) applies the brakes Shade the area on the graph that represents the distance travelled by the car during the period.
Which part of the graph represents uniform motion of the car?
In Fig. 9.26, A, B, C and D are four points on a circle. AC and BD intersect at a point E such that ∠ BEC = 130° and ∠ ECD = 20°. Find ∠ BAC.