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Question 0101c
Equations and Inequalities
Inequalities III
Question
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What is the approach to solving exactly the inequality
x
2
−
5
x
+
26
x
+
3
>
0
?
\frac{x^2 - 5 x + 26}{x + 3} > 0?
x
+
3
x
2
−
5
x
+
26
>
0
?
Attempt
Select the correct option:
Factorize
x
2
−
5
x
+
26
{x^2 - 5 x + 26}
x
2
−
5
x
+
26
into the form
(
x
+
p
)
(
x
+
q
)
{(x+p)(x+q)}
(
x
+
p
)
(
x
+
q
)
to find the critical points.
"Cross multiply" by multiplying
x
+
3
{x + 3}
x
+
3
to the RHS of the equation.
Use the quadratic formula
x
=
−
b
±
b
2
−
4
a
c
2
a
{x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}}
x
=
2
a
−
b
±
b
2
−
4
a
c
for
x
2
−
5
x
+
26
=
0
{x^2 - 5 x + 26=0}
x
2
−
5
x
+
26
=
0
to find the critical points.
Complete the square for
x
2
−
5
x
+
26
{x^2 - 5 x + 26}
x
2
−
5
x
+
26
into the form
(
x
+
p
)
2
+
q
{(x+p)^2 + q}
(
x
+
p
)
2
+
q
to show that it is always positive.
Answer
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