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Question 0107a
Equations and Inequalities
2007 Paper 1 Question 1 Variant
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Show that
4
x
+
9
x
2
+
4
x
+
3
+
1
=
x
2
+
8
x
+
12
x
2
+
4
x
+
3
.
\frac{ 4 x + 9 }{ x^2 + 4 x + 3 } + 1=\frac{ x^2 + 8 x + 12 }{ x^2 + 4 x + 3 }.
x
2
+
4
x
+
3
4
x
+
9
+
1
=
x
2
+
4
x
+
3
x
2
+
8
x
+
12
.
Hence, without using a calculator, solve the inequality
4
x
+
9
x
2
+
4
x
+
3
<
−
1
\frac{ 4 x + 9 }{ x^2 + 4 x + 3 } < -1
x
2
+
4
x
+
3
4
x
+
9
<
−
1
[5]
Answer
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