Question

Find the roots of each of the following quadratic equations if they exist by the method of completing the squares:
2x2 – 5x + 3 = 0

Answer 1

Question:

Find the roots of each of the following quadratic equations if they exist by the method of completing the squares:

2x² – 5x + 3 = 0

Answer:

x = 1 , 3/2

Note:

• An equation of degree 2 is know as quadratic equation .

• Roots of an equation is defined as the possible values of the unknown (variable) for which the equation is satisfied.

• The maximum number of roots of an equation will be equal to its degree.

• A quadratic equation has atmost two roots.

• The general form of a quadratic equation is given as , ax² + bx + c = 0 .

• The discriminant of the quadratic equation is given as , D = b² - 4ac .

• If D = 0 , then the quadratic equation would have real and equal roots .

• The discriminant of the quadratic equation is given as , D = b² - 4ac .

• If D = 0 , then the quadratic equation would have real and equal roots .

• If D > 0 , then the quadratic equation would have real and distinct roots .

• If D < 0 , then the quadratic equation would have imaginary roots

Solution:

Here,

The given quadratic equation is :

2x² – 5x + 3 = 0

Clearly,

Here we have ;

a = 2

b = -5

c = 3

Thus,

The determinant of the given quadratic equation will be given as ;

=> D = b² - 4ac

=> D = (-5)² - 4•2•3

=> D = 25 - 24

=> D = 1 {D ≥ 0}

Clearly,

The determinant of the given quadratic equation is greater than zero.

Thus, there must exist real roots of the given quadratic equation.

Now,

=> 2x² – 5x + 3 = 0

=> x² - 5x/2 + 3/2 = 0

=> x² - 5x/2 + (5/4)² - (5/4)² + 3/2 = 0

=> x² - 2•x•(5/4) + (5/4)² = (5/4)² - 3/2

=> (x - 5/4)² = 25/16 - 3/2

=> (x - 5/4)² = (25-24)/16

=> (x - 5/4)² = 1/16

=> x - 5/4 = √(1/16)

=> x - 5/4 = ± 1/4

=> x = 5/4 ± 1/4

If x = 5/4 + 1/4

=> x = (5+1)/4

=> x = 6/4

=> x = 3/2

If x = 5/4 - 1/4

=> x = (5-1)/4

=> x = 4/4

=> x = 1

Hence,

The roots of the given quadratic equation are :

x = 1 , 3/2

or

x = 1 , 1.5

Answer 2

$\huge{\boxed{\purple{Answer}}}$

Given Equation :- $2x^{2}-5x+3=0$

Method to solve :- Completing squares method

$\large{\boxed{\green{Solution}}}$

First let's verify if the given equation has roots or not

To verify that the equation has got roots , we need to calculate the value of the discriminent

$\large{\underline{\red{Discriminent}}}$

• If $ax^{2}+bx+c$ is a quadratic equation, then the discriminent is given by,
• D = $\sqrt{b^{2}-4ac}$
• If , D > 0 , then two distinct roots exist
• If , D = 0 , then two real and repeated roots exist
• If , D < 0 , then no real roots exist

Now lets verify the discriminent of the given equation,

In the given equation , the values are as follows ,

• a = 2
• b = -5
• c = 3
• D = $\sqrt{b^{2}-4ac}$

• D = $\sqrt{{-5}^{2}-4(2)(3)}$

• D = $\sqrt{25-24}$

• D = $\sqrt{1}$

• D = 1

Here D > 0 , Therefore, roots are real and distinct

Now lets find out the roots by completing square method

$\large{\underline{\pink{Completing\;square\;method}}}$

Given Equation :- $2x^{2}-5x+3=0$

Dividing the equation by 'a' coefficient of ${x}^{2}$

$=>\dfrac{2{x}^{2}}{2}-\dfrac{5x}{2}+\dfrac{3}{2}$

$=>\dfrac{2{x}^{2}}{2}-\dfrac{5x}{2}=-\dfrac{3}{2}$

$=>x^{2}-\dfrac{5x}{2}=-\dfrac{3}{2}$

Multiplying and dividing by '2' on for second term on L.H.S

$=>x^{2}-\dfrac{2}{2}(\dfrac{5x}{2})=-\dfrac{3}{2}$

Adding ${(\dfrac{5}{4})}^{2}$ on both sides

$=>x^{2}-2(\dfrac{5x}{4})+{(\dfrac{5}{4})}^{2}=-\dfrac{3}{2}+{(\dfrac{5}{4})}^{2}$

L.H.S is in the form of ${(a-b)}^{2}={a}^{2}+{b}^{2}-2ab$

So , now by simplifying,

$=>{x-\dfrac{5}{4}}^{2}=\dfrac{1}{16}$

$=>x-\dfrac{5}{4}=+\;or\;-\dfrac{1}{4}$

$=>x=\dfrac{5}{4}+\;or\;-\dfrac{1}{4}$

$=>x=\dfrac{6}{4}\;or\;\dfrac{4}{4}$

$=>x=1.5\;or\;1$

$\boxed{\green{ The\;required\;values\;of\;x\;are\;1\;and\;1.5}}$