sqr(3-x)-sqr(1+x) >1/2.......................find x

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

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We'll impose constraints of existence of square roots:

3 - x >= 0

x =< 3

1 + x >= 0

x >= -1

The range of admissible values for x is [-1 ; 3].

Now, we'll solving the inequality by raising to square both sides:

3 - x + 1 + x - 2sqr(3-x)*(1+x) > 1/4

We'll eliminate x and we'll combine like terms:

4 - 2sqr(3-x)*(1+x) > 1/4

We'll subtract 4 both sides:

- 2sqr(3-x)*(1+x) > 1/4 - 4

- 2sqr(3-x)*(1+x) > -15/4

We'll multiply by -1:

2sqr(3-x)*(1+x) < 15/4

We'll divide by 2:

sqr(3-x)*(1+x) < 15/8

We'll raise to square again:

(3-x)*(1+x) < 225/64

We'll subtract 225/64:

(3-x)*(1+x) - 225/64 < 0

We'll remove the brackets:

3 + 2x - x^2 - 225/64 < 0

64*3 + 64*2x - 64x^2 - 225 < 0

- 64x^2 + 128x - 33 < 0

64x^2 - 128x + 33 > 0

The expresison above is strictly positive if discriminant of the quadratic is negative:

delta = 16384 - 8448

delta = 7936

Since delta >0, the quadratic has 2 real roots.

x1 = (128+16sqrt31)/2*64

x1 = 16(8+sqrt31)/2*2*2*16

x1 = (8+sqrt31)/8

x2 = (8-sqrt31)/8

The expression is positive over the ranges [-1;(8-sqrt31)/8) U ((8+sqrt31)/8 ; 3].

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