Question

Find the roots of the following quadratic equation if they exist by the method of completing square 2x squared minus 7x + 3 equal to 0

Answer 1

Answer:

x=√43+7/2 or -√43+7/2

Step-by-step explanation:

2x^2-7x+3=0

2x^2-7=-3

divide both sides by 2

x^2-7/2=-3/2

3rd term=( 1/2*-7/2)^2

=( -7/2)^2

=49/4

add 49/4 on both sides

x^2-7/2+49/4=-3/2+49/4

x^2-7/2+(7/2)^2=-6/49/4

(x^2-7/2)^2=43/4

taking square root

x-7/2=+/-√43/2

therefore x=√43+7/2 or -√43+7/2

Answer 2

${\bold{\blue{\underline{\red{S}\pink{ol}\green{uti}\purple{on:-}}}}}$

⤤ Given quadratic equation,

• 2x² - 7x + 3 = 0

To find :-

• Roots of the quadratic equation

Solution :-

• a = 2 , b = -7 and c = 3

So,

D = b² - 4ac

➝ D = (-7)² - 4 × 2 × 3

➝ D = 49 - 24

➝ D = 25

So,

D > 0

Then,

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

$\rm\longrightarrow\:2{x}^{2}-7x=-3$

⤤ Dividing both side by the value of a

$\rm\longrightarrow\:{x}^{2}-\dfrac{7}{2}x=\dfrac{-3}{2}$

$\rm\star\:Adding\:\bigg(\dfrac{b}{2a}\bigg)^2\:on\:both\:the\:sides$

$\rm\:{x}^{2}-\dfrac{7}{2}x+\bigg(\dfrac{7}{4}\bigg)^2=\dfrac{-3}{2}+\bigg(\dfrac{7}{4}\bigg)^2$

$\rm\longrightarrow\:{x}^{2}-\dfrac{7}{2}+\dfrac{49}{16}=\dfrac{-3}{2}+\dfrac{49}{16}$

$\rm\longrightarrow\:({x})^{2}-2\times\:x\times\:\dfrac{7}{4}+\bigg(\dfrac{7}{4}\bigg)^2=\dfrac{-24+49}{16}$

$\rm\longrightarrow\:\bigg(x-\dfrac{7}{4}\bigg)^2=\dfrac{25}{16}$

$\rm\longrightarrow\:x-\dfrac{7}{4}=\pm\sqrt{\dfrac{25}{16}}$

$\rm\longrightarrow\:x-\dfrac{7}{4}=\pm\dfrac{5}{4}$

$\rm\longrightarrow\:x-\dfrac{7}{4}=\dfrac{5}{4}$

$\rm\longrightarrow\:x=\dfrac{5}{4}+\dfrac{7}{4}$

$\rm\longrightarrow\:x=\dfrac{5+7}{4}$

$\rm\longrightarrow\:x=\cancel\dfrac{12}{4}$

$\rm\longrightarrow\:x=3$

Then,

$\rm\longrightarrow\:x-\dfrac{7}{4}=\dfrac{-5}{4}$

$\rm\longrightarrow\:x=\dfrac{-5}{4}+\dfrac{7}{4}$

$\rm\longrightarrow\:x=\dfrac{-5+7}{4}$

$\rm\longrightarrow\:x=\dfrac{\cancel{2}}{\cancel{4}}$

$\rm\longrightarrow\:x=\dfrac{1}{2}$

Hence,the roots of the quadratic equation will be 3 and ½.