Question

FIND ROOTS OF FOLLOWING QUADRATIC EQUATION IF THEY EXIST BY COMPLETING THE SQUARE METHOD
$2x^{2} - 7x + 3 = 0$


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Answer 1

Step-by-step explanation:

b²-4ac = -7² - 4 (2) 3

= 49 - 24

= 25

by quadratic formula

-(-7) + 5/ 4 = 7 + 5 /4 = 3

or 7 -5/4 = 2/ 4 or 1/2

Answer 2

Step-by-step explanation:

Given:-

2x² - 7x + 3 = 0

Solution:-

All Equation of the form of ax² + bx + c can should using the quadratic formula:

$\boxed{ \rm \mathrm{\frac{ - b± \sqrt{ {b}^{2} - 4ac} }{2a} }}$

The quadratic formula gives two solution one when ± is adding and one when it is subtracting.

2x² - 7x + 3 = 0

This equation is in standard form ax + bx + c is = 0. Substituting 2 for a, -7 for b, and 3 for c in the quadratic formula,

$\boxed{ \rm \mathrm{\frac{ - b± \sqrt{ {b}^{2} - 4ac} }{2a} }}$

$\rm \: x = \mathrm{\frac{ -( - 7)± \sqrt{( { - 7})^{2} - 4 \times 2 \times 3} }{2 \times 2} }$

Square -7.

$\rm \: x = \mathrm{\frac{ -( - 7)± \sqrt{49 - 4 \times 2 \times 3} }{2 \times 2} }$

Multiply -4 times 2.

$\rm \: x = \mathrm{\frac{ -( - 7)± \sqrt{49 - 8 \times 3} }{2 \times 2} }$

$\rm \: x = \mathrm{\frac{ -( - 7)± \sqrt{49 - 24} }{2 \times 2} }$

Add 49 to -24.

$\rm \: x = \mathrm{\frac{ -( - 7)± \sqrt{25} }{2 \times 2} }$

Take the square root of 25.

$\rm \: x = \mathrm{\frac{ -( - 7)± 5 }{2 \times 2} }$

The opposite of -7 is 7.

$\rm \: x = \mathrm{\frac{ 7± 5 }{2 \times 2} }$

Multiply the denominator 2 times 2.

$\rm \: x = \mathrm{\frac{ 7± 5 }{4} }$

Now, solve the equation $\rm \: x = \mathrm{\frac{ 7± 5 }{4} }$ when ± is plus. Add 7 to 5.

$\rm \: x = \frac{12}{4}$

Divide 12 by 4.

$\rm \: x = 4$

Now, solve the equation $\rm \: x = \mathrm{\frac{ 7± 5 }{4} }$ when ± is minus. And subtract 5 from 7.

$\rm \: x = \frac{2}{4}$

Reduce the fraction 2/4 to lowest by extracting and cancelling out 2.

The equation is now solved

x = 3, x = 1/2

Answer:-

x = 1/2 = 0.5

x = 3