Question

Solve 2x2 - 7x + 3 = 0 by completing square method​

Answer 1

Answer:

let 2x2 -7x+k=(x-a)2

2x2-7x+k=x2-2ax+a2

comparing coefficients of x and constants on both sides

2a=7

a=7/2

and

k=a2

k=7/22

k=49/4

now,

(2x2-7x+49/4)-49/4+3=0

3x2-7x+49/4=49+12/4

(x-7/2)2=61/4

(x-7/2)2=root61/4

(x-7/2)2=root 61/2

(x-7/2)=+-root61/2...(Taking square on both sides)

x-7/2=root 61/2

or x-7/2 =root61/2

x=7-root 61/2

or x=7-root 61/2

Step-by-step explanation:

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Answer 2

AnswEr:-

❀ Your Answer is 1/2 And 3.

ExplanaTion:-

Given equation:-

• $\tt\leadsto 2x^2-7x+3.$

To Find:-

• The Value of x by Completing square method.

ID Used:-

$\\ \tt\longrightarrow(a - b) {}^{2} = {a}^{2} - 2ab + {b}^{2} .$

So lets solve the given equation:-

$\\ \\ \tt\mapsto2 {x}^{2} - 7x + 3 = 0. \\ \\ \tt\mapsto \dfrac{\cancel2 {x}^{2} }{\cancel2} - \frac{7x}{2} + \frac{3}{2} = \frac{0}{2} . \\ \\ \rm(divding \: \: the \: \: whole \: \: equation \: \: by \: \: 2). \\ \\ \tt\mapsto x {}^{2} - \frac{7x}{2} + \frac{3}{2} =0. \\ \\ \tt\mapsto {x}^{2} - \frac{7x}{2} = - \frac{3}{2} . \\ \\ \tt\mapsto {x}^{2} -( 2 \times x \times \frac{7}{4} ) = - \frac{ 3}{2} . \\ \\ \tt\mapsto {(x)}^{2} - (2 \times x \times \frac{7}{4} ) + ( \frac{7}{4} ) {}^{2} = - \frac{3}{2} + ( \frac{7}{4} ) {}^{2} . \\ \\ \rm(adding \: ( \frac{7}{4}) {}^{2} \: \: to \: \: both \: \: sides) . \\ \\ \tt\mapsto(x - \frac{7}{4} ) {}^{2} = ( \frac{7}{4}) {}^{2} - \frac{3}{2}. \\ \\ \tt\mapsto(x - \frac{7}{4} ) {}^{2} = \frac{49}{16} - \frac{3}{2} . \\ \\ \tt\mapsto(x - \frac{7}{4} ) {}^{2} = \frac{49 - 24}{16} .\\ \\ \rm(by \: \: taking \: \: lcm). \\ \\ \tt\mapsto(x - \frac{7}{4} ) {}^{2} = \frac{25}{36} . \\ \\ \tt\mapsto(x - \frac{7}{4} ) {}^{\cancel2} = \pm(\frac{5}{6} ){}^{\cancel2}. \\ \\ \tt\mapsto x - \frac{7}{4} = \pm\frac{5}{6} . \\ \\\tt\mapsto x = \pm \frac{5}{4} + \frac{7}{4} . \\ \\ \tt\mapsto x = \frac{7 - 5}{4} \: \: \: \: or \: \: \: \: x = \frac{7 + 5}{4} .\\ \\ \tt\mapsto x = \frac{2}{4} \: \: \: \: or \: \: \: \: x = \frac{12}{4} . \\ \\ \tt\mapsto\underline{\underline {x = \frac{1}{2} \: \: \: \: or \: \: \: \: x = 3.}}$

Therefore the required values of x are $\tt\dfrac{1}{2}$ And 3.