Question

find the root of following quadratic equation by method of completing square 2x2-7x+3=0
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Answer 1

Answer:

$\boxed{\textbf{x = 3, 1/2}}$

Step-by-step explanation:

$given \ quadratic \ equation, \\ 2x^2-7x+3=0 \\\\ divide \ by \ 2 \ on \ both \ sides, \\\\ \frac{1}{2}(2x^2-7x+3) = \frac{1}{2}(0) \\\\ x^2-\frac{7}{2}x +\frac{3}{2} =0 \\\\x^2-\frac{7}{2}x=-\frac{3}{2} \\\\ x^2-2(x)(\frac{7}{4}) = -\frac{3}{2} \\\\ adding \ (\frac{7}{4})^2 \ on \ both \ sides \\\\x^2-2(x)(\frac{7}{4}) + (\frac{7}{4})^2 = -\frac{3}{2} + (\frac{7}{4})^2$

$[a^2-2ab+b^2=(a-b)^2] \\\\ (x-\frac{7}{4})^2= -\frac{3}{2}+\frac{49}{16} \\\\ (x-\frac{7}{4})^2= \frac{49-24}{16} \\\\ (x-\frac{7}{4})^2=\frac{25}{16} \\\\ (x-\frac{7}{4})= \pm \sqrt{\frac{25}{16}} \\\\ (x-\frac{7}{4})=\pm{\frac{5}{4}} \\\\ x=\frac{7}{4} \pm\frac{5}{4} \\\\ => x=\frac{7}{4}+\frac{5}{4} =\frac{12}{4}=3 \\\\ =>x=\frac{7}{4}-\frac{5}{4} =\frac{2}{4} = \frac{1}{2} \\\\ \textbf{x = 3, 1/2}$

hope it helps..!

Answer 2

2x2 – 7x + 3 = 0  ⇒ 2x2 – 7x = – 3

Dividing by 2 on both sides, we get

⇒ x2 -7x/2 = -3/2

⇒ x2 -2 × x ×7/4 = -3/2

On adding (7/4)2 to both sides of equation, we get

⇒ (x)2-2×x×7/4 +(7/4)2 = (7/4)2-3/2

⇒ (x-7/4)2 = (49/16) – (3/2)

⇒(x-7/4)2 = 25/16  

⇒(x-7/4)2 = ±5/4

⇒ x = 7/4 ± 5/4

⇒ x = 7/4 + 5/4 or x = 7/4 – 5/4

⇒ x = 12/4 or x = 2/4

⇒ x = 3 or x = 1/2