.When we divide the sum of −13 5 and 12 7 by the product of −31 7 and −1 2 ,we get

Question

.When we divide the sum of −13
5
and 12
7
by the product of −31
7
and −1
2
,we get
(a) -2/5 (b) 4/7 (c) 13/2 (d) -12/7

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Aubrey 1 week 2021-09-13T14:25:50+00:00 1 Answer 0 views 0

Answers ( )

    0
    2021-09-13T14:26:52+00:00

    Step-by-step explanation:

    (i) 2×2–7x+ 3 = 0

    On dividing both sides of the equation by 2, we get

    x2– 7x/2 +3/2=0

    x2– 2 ×x× 7/4 +3/2=0

    adding (7/4)2 and subtracting on LHS, we get

    (x)2- 2 ×x× 7/4 + (7/4)2- (7/4)2+ 3/2=0

    (x- 7/4)2= 49/16 – 3/2

    (x- 7/4)2= 25/16

    (x- 7/4) =± 5/4

    x= 7/4± 5/4

    x= 7/4+ 5/4 orx= 7/4 – 5/4

    x= 12/4 orx= 2/4

    x= 3 or 1/2

    (ii) 2×2+x– 4 = 0

    On dividing both sides of the equation, we get

    x2+x/2 – 2=0

    adding (1/4)2and Subtracting to LHS, we get

    (x)2+2 ×x× 1/4 + (1/4)2- (1/4)2 -2 =0

    (x+ 1/4)2= 33/16

    ⇒x+ 1/4 = ± √33/4

    ⇒x= ± √33/4 – 1/4

    ⇒x= ± √33-1/4

    ⇒x= √33-1/4 orx= -√33-1/4

    (iii) 4×2+ 4√3x+ 3 = 0

    (2x)2+ 2 × 2x× √3+ (√3)2= 0

    (2x+ √3)2= 0

    (2x+ √3) = 0 and (2x+ √3) = 0

    x= -√3/2 orx= -√3/2

    (iv) 2×2+x+ 4 = 0

    On dividing both sides of the equation, we get

    x2+ 1/2x+2=0

    Adding and Subtracting (1/4)2to LHS, we get

    (x)2+2 ×x× 1/4 + (1/4)2- (1/4)2+ 2=0

    (x+ 1/4)2= 1/16 – 2

    (x+ 1/4)2= -31/16

    However, the square of number cannot be negative.

    Therefore, there is no real root for the given equation

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