Question

Solve 2x^2-18x+40=0 by completing the square method?​

Answer 1

Answer:

2x2-18x-40=0

Two solutions were found :

x =(9-√161)/2=-1.844

x =(9+√161)/2=10.844

Step by step solution :

Step 1 :

Equation at the end of step 1 :

(2x2 - 18x) - 40 = 0

Step 2 :

Step 3 :

Pulling out like terms :

3.1 Pull out like factors :

2x2 - 18x - 40 = 2 • (x2 - 9x - 20)

Trying to factor by splitting the middle term

3.2 Factoring x2 - 9x - 20

The first term is, x2 its coefficient is 1 .

The middle term is, -9x its coefficient is -9 .

The last term, "the constant", is -20

Step-1 : Multiply the coefficient of the first term by the constant 1 • -20 = -20

Step-2 : Find two factors of -20 whose sum equals the coefficient of the middle term, which is -9 .

-20 + 1 = -19

-10 + 2 = -8

-5 + 4 = -1

-4 + 5 = 1

-2 + 10 = 8

-1 + 20 = 19

Observation : No two such factors can be found !!

Conclusion : Trinomial can not be factored

Equation at the end of step 3 :

2 • (x2 - 9x - 20) = 0

Step 4 :

Equations which are never true :

4.1 Solve : 2 = 0

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Answer 2

$\huge{ \boxed{ \bigstar{ \purple{ \underline{ \mathbf{♡ANSWER♡}}}}}}</p><p>$

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${ \boxed{ \green{ \underline{ \mathbf{by \: compleating \: the \: square \: method}}}}} = > </p><p>$

${ \boxed{ \red{ \underline{ \mathbf{equation = 2 {x}^{2} - 18x + 40}}}}} </p><p>$

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$\huge{ \boxed{ \blue{ \underline{ \mathbf{explantion}}}}} = > </p><p>$

Divide both side by 2,

we get =>

$\huge {x}^{2} - 9x + 20 = 0$

$\huge{add{ (\frac{9}{2 })^{2} }on \: both \: side }$

we get,=>

$\large{x - 9x + \frac{81}{4} + 20 = \frac{81}{4} }$

$\large(x - { \frac{9}{2}) }^{2} + 20 = \frac{81}{4} \: \\ \\ {(x - \frac{9}{2} ) }^{2} = \frac{81}{4 } - 20 \\ \\ {(x - \frac{9}{2} ) }^{2} = \frac{81 - 80}{4} \\ \\ {(x - \frac{9}{2} )}^{2} = \frac{1}{4} \\ \\ x - \frac{9}{2 } = \sqrt{ \frac{1}{4} } \\ \\ x - \frac{9}{2} = \frac{1}{2} \\ \\ x = \frac{1}{2} + \frac{9}{2} \\ \\ x = \frac{10}{2} \\ \\ x = 5$

$\huge{or}$

$\large x = \frac{ - 1}{2} + \frac{9}{2} \\ \\ x = \frac{8}{2} \\ \\ x = 4$

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hops this may help you

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