Find the roots of the following quadratic equation if they exist by the method of completing the square 2x²-7x+3=0?

Jun 23, 2018

See my proof below.

Explanation:

${x}^{2} - 2 \cdot \frac{7}{4} x + \frac{49}{16} + \frac{3}{2} - \frac{49}{16} = 0$
this is

${\left(x - \frac{7}{4}\right)}^{2} - \frac{25}{16} = 0$
using that

(a^2-b^2=(a-b)(a+b)
then we get

$\left(x - \frac{7}{4} + \frac{5}{4}\right) \left(x - \frac{7}{4} - \frac{5}{4}\right) = 0$

$\left(x - \frac{1}{2}\right) \left(x - 3\right) = 0$

Jun 23, 2018

$x = \frac{1}{2}$ and $x = 3$

Explanation:

It is always good to write out the square as a reminder. We want the expression on the form:
${\left(x - a\right)}^{2} = b$
$2 {\left(x - a\right)}^{2} = 2 \left({x}^{2} - 2 a x + {a}^{2}\right) = 2 {x}^{2} - 7 x = - 3$
gives
${\left(x - a\right)}^{2} = \left({x}^{2} - 2 a x + {a}^{2}\right) = {x}^{2} - \frac{7}{2} x + {\left(\frac{7}{4}\right)}^{2}$
=$- \frac{3}{2} + {\left(\frac{7}{4}\right)}^{2} = \frac{- 24 + 49}{16} = \frac{25}{16}$
where I have moved the constant over to the right side.

Comparing term for term we see that
$2 a = \frac{7}{2}$ or $a = \frac{7}{4}$
The square, therefore, is
${\left(x - \frac{7}{4}\right)}^{2} = \frac{25}{16}$

We, therefore get:
=$x - \frac{7}{4} = \pm \frac{5}{4}$

Therefore:
$x = \frac{7}{4} \pm \frac{5}{4}$
This gives the two solutions $x = \frac{1}{2}$ and $x = 3$

Test:

We see the graph of $f \left(x\right) = 2 {x}^{2} - 7 x + 3$ crosses the x axis in x=0.5 and x=3