Find the range of values of x for which $(2x-1)^2 < (x+3)^2$ by algebraic method.
Answer (2)
The range of values of x for which ( 2 x − 1 ) 2 < ( x + 3 ) 2 is − 3 2 < x < 4 .
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To find the range of values of x for which ( 2 x − 1 ) 2 < ( x + 3 ) 2 , we can solve this step by step using algebra.
Expand both sides of the inequality:
( 2 x − 1 ) 2 = ( 2 x − 1 ) ( 2 x − 1 ) = 4 x 2 − 4 x + 1
( x + 3 ) 2 = ( x + 3 ) ( x + 3 ) = x 2 + 6 x + 9
Set up the inequality:
4 x 2 − 4 x + 1 < x 2 + 6 x + 9
Move all terms to one side to form a quadratic inequality:
4 x 2 − 4 x + 1 − x 2 − 6 x − 9 < 0
Simplifying gives:
3 x 2 − 10 x − 8 < 0
Factor the quadratic expression:
The quadratic expression 3 x 2 − 10 x − 8 can be factored to:
( 3 x + 2 ) ( x − 4 ) < 0
To find the critical points, set each factor to zero:
3 x + 2 = 0 ⇒ x = − 3 2 x − 4 = 0 ⇒ x = 4
Test intervals determined by the critical points:
The critical points split the real number line into intervals: ( − ∞ , − 3 2 ) , ( − 3 2 , 4 ) , ( 4 , ∞ ) .
Choose a test point in each interval to determine where the inequality holds:
For x = − 1 (in ( − ∞ , − 3 2 ) ):
( 3 ( − 1 ) + 2 ) (( − 1 ) − 4 ) = ( − 3 + 2 ) ( − 5 ) = − 1 ( − 5 ) = 5 (positive)
For x = 0 (in ( − 3 2 , 4 ) ):
( 3 ( 0 ) + 2 ) (( 0 ) − 4 ) = 2 ( − 4 ) = − 8 (negative)
For x = 5 (in ( 4 , ∞ ) ):
( 3 ( 5 ) + 2 ) (( 5 ) − 4 ) = ( 15 + 2 ) ( 1 ) = 17 (positive)
Determine the solution interval:
The inequality ( 3 x + 2 ) ( x − 4 ) < 0 holds in the interval where the test point yielded a negative product, which is:
(
(-\frac{2}{3}, 4))
Therefore, the solution to the inequality ( 2 x − 1 ) 2 < ( x + 3 ) 2 is − 3 2 < x < 4 . This means x is any value in the range ( − 3 2 , 4 ) .
