# How do you solve 4x^2 - 12x + 9 = 0?

Jun 23, 2016

$x = \frac{3}{2}$

#### Explanation:

In $4 {x}^{2} - 12 x + 9 = 0$ as niddle term $- 12 x$ is twice the product (with negative sign) of square roots of first ($s q t \left(4 {x}^{2}\right) = 2 x$) and third term ($\sqrt{9} = 3$,

it can be factorized using formula ${a}^{2} - 2 a b + {b}^{2} = {\left(a - b\right)}^{2}$

Hence $4 {x}^{2} - 12 x + 9 = 0$

$\Leftrightarrow {\left(2 x\right)}^{2} - 2 \times 3 \times 3 x + {3}^{2} = 0$

or ${\left(2 x - 3\right)}^{2} = 0$

and hence $2 x = 3$ or $x = \frac{3}{2}$
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