Quadratics ineqs-problems!?

Question

1)Find the range of values of x for which:

x^2-7x+10/x^2+3<=0

2)Find the range of values of x for which:

3/x+1<=0

Answer 1

this equation can be simplified as

if you cud have placed the brackets properly it wud be easy to understand the question

well i can solve 2nd part if there are no brackets

3/x+1=(3+x)/x&lt;=0

multiplying both sides by x^2 as it is alaways positive it wont matter

now the equality is

(3+x)(x)&lt;=0

by wavy curve method

we get interval [-3,0]

Answer 2

1)

we know x^2 + 3 is positive

so x^2 - 7x + 10 &lt;= 0

or (x-2)(x-5) &lt;=0

x &lt; 2 means both are -ve so product positive

2 &lt;= x &lt;= 5 condition meets

x &gt; 5 means both are +ve so product positive

so 2 &lt;= x &lt;= 5

2nd one

3/(x+1) &lt;= 0

means x + 1 &lt; 0

or x &lt; - 1

math kp

Answer 3

1))since the denominator is always  positive

check for which x numerator is negetive

x^2-7x+10=(x-5)(x-2)

hence 1) is true whn x belongs to the set (-infinity,2]U[5,infinity)

2))  (3/x)   +1&lt;=0

3/x&lt;=-1

x/3&gt;=-1

x&gt;=-3   and x must be &lt; 0