How do you solve $(4x – 3) ² – (2x – 1) ² = 0$ ?

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Answer 1 · 2016-05-15T17:32:18

$x=2/3, 1$

The given expression i of the form $a^2-b^2$

Where $a=4x-3$ and $b=2x-1$

Now, $color(red)(a^2-b^2 = (a+b)(a-b))$

Then,
$(4x-3)^2-(2x-1)^2 =0$

$=> (4x-3+2x-1){4x-3-(2x-1)}=0$

$=> (6x-4)(4x-3-2x+1)=0$

$=> (6x-4)(2x-2)=0$

$color(red)( 6x-4=0)$ or $color(blue)(2x-2=0)$

$color(red)( 6x=4)$ or $color(blue)(2x=2)$

$color(red)( x=4/6)$ or $color(blue)(x=2/2)$

$color(red)( x=2/3)$ or $color(blue)(x=1)$

How do you solve (4x – 3) ² – (2x – 1) ² = 0(4x – 3) ² – (2x – 1) ² = 0?