Question

x
${x}^{2} - 7x + 3 = 0$
-7x+3=0

​

Answer 1

Answer:

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Answer 2

$\huge{❥}\:{\mathtt{{\purple{\boxed{\tt{\pink{\red{A}\pink{n}\orange{s}\green{w}\blue{e}\purple{r᭄}}}}}}}}❥$

$\large \bold{ {x}^{2} - 7x + 3 = 0 }$

Deviding by 2

$\bold{\frac{2x^{2} - 7x + 3}{2} = \frac{0}{2} }$

$\bold{\frac{ {2x}^{2} }{2} - \frac{7x}{2} - \frac{3}{2} = \frac{0}{2}}$

$\bold{ \frac{ \cancel{2}x^{2} }{ \cancel{2}} - \frac{7x}{2} + \frac{3}{2} = \frac{0}{2} }$

$\bold{ {x}^{2} - \frac{7x}{2} + \frac{3}{2} = 0}$

We know that,

$\bold{ {(a + b)}^{2} = {a}^{2} - 2ab + {b}^{2} }$

Here a = x and

$\bold{ - 2ab = \frac{ - 7x}{2} }$

$\bold{ - 2xb = \frac{ - 7x}{2} }$

$\bold{b = \frac{ - 7x}{2( - 2x)} }$

$\bold{b = \frac{7}{4} }$

Now, in our equation is

$\bold{ {x}^{2} - \frac{7x}{2} + \frac{3}{2} } = 0$

$\bold{Adding \: And \: \: Subtracting} ( \frac{7}{4} )^{2}$

$\bold{x^{2} - \frac{7x}{2} + \frac{3}{2} } + \bold{( \frac{7}{4} )^{2} - ( \frac{7}{4} )^{2} = 0}$

$\bold{ {x}^{2} + ( \frac{7}{4} )^{2} - \frac{7x}{2} + \frac{3}{2} - ( \frac{7}{4})^{2} = 0 }$

$\bold{(x - \frac{7}{4})^{2} + \frac{3}{2} - ( \frac{7}{4} )^{2} = 0 }$

$\bold{(x - \frac{7}{4})^{2} + \frac{3}{2} - \frac{49}{16} = 0 }$

$\bold{(x - \frac{7}{4})^{2} + \frac{3(8) - 49}{16} = 0}$

$\bold{(x - \frac{7}{4})^{2} + \frac{24- 49}{16} = 0}$

$\bold{(x - \frac{7}{4})^{2} - \frac{25}{16} = 0}$

$\bold{(x - \frac{7}{4})^{2} = \frac{25}{16} }$

$\bold{(x - \frac{7}{4})^{2} = ( \frac{5}{4})^{2} }$

$\bold{x - \frac{7}{4} = ± \frac{5}{4} }$

$\bold{x - \frac{7}{4} = \frac{5}{4} }$

$\bold{x = \frac{5}{4} + \frac{7}{4} }$

$\bold{x = \frac{5 + 7}{4} }$

$\bold{x = \frac{ \cancel{12}}{ \cancel{4}} }$

$\bold{x = 3}$

$\bold{x - \frac{7}{4} = \frac{ - 5}{4} }$

$\bold{x = \frac{ - 5}{4} + \frac{7}{4} }$

$\bold{ x = \frac{ - 5 + 7}{4} }$

$\bold{x = \frac{ \cancel2}{ \cancel4} }$

$\bold{x = \frac{1}{2} }$

HENCE,

$\huge \bold{x = 3 \: and \: x = \frac{1}{2} }$

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