Question

For 2x2— 5x —3 = 0, determine which of the following are solutions?
(i) x = 3 (ii) x= –2
(iii) x=-1/2 (iv) x=-1/3

Answer 1

Note:

• The possible values of unknown (variable) for which the equation is satisfied are called its solutions or roots .

• If x = a is a solution of any equation in x , then it must satisfy the given equation otherwise it's not a solution (root) of the equation.

Solution:

Here,

The given equation is : 2x² - 5x - 3 = 0 ------(1)

(I) Let's check whether x = 3 is a solution of eq-(1) .

Putting x = 3 in eq-(1) , we have ;

=> 2•(3)² - 5•3 - 3 = 0

=> 18 - 15 - 3 = 0

=> 0 = 0 (which is true)

Since , eq-(1) is satisfied by x = 3 , thus x = 3 is a solution of eq-(1) .

(ii) Let's check whether x = -2 is a solution of eq-(1) .

Putting x = -2 in eq-(1) , we have ;

=> 2•(-2)² - 5•(-2) - 3 = 0

=> 8 + 10 - 3 = 0

=> 15 = 0 (which is not true)

Since , eq-(1) is not satisfied by x = -2 , thus x = -2 is not a solution of eq-(1) .

(iii) Let's check whether x = -1/2 is a solution of eq-(1) .

Putting x = -1/2 in eq-(1) , we have ;

=> 2•(-1/2)² - 5•(-1/2) - 3 = 0

=> 1/2 + 5/2 - 3 = 0

=> 6/2 - 3 = 0

=> 3 - 3 = 0

=> 0 = 0 (which is true)

Since , eq-(1) is satisfied by x = -1/2 , thus x = -1/2 is a solution of eq-(1) .

(iv) Let's check whether x = -1/3 is a solution of eq-(1) .

Putting x = -1/3 in eq-(1) , we have ;

=> 2•(-1/3)² - 5•(-1/3) - 3 = 0

=> 2/9 + 5/3 - 3 = 0

=> (2 + 15 - 27)/3 = 0

=> -7/3 = 0 (which is not true)

Since , eq-(1) is not satisfied by x = -1/3 , thus x = -1/3 is not a solution of eq-(1) .

By factorisation :–

The given quadratic equation is:

=> 2x² - 5x - 3 = 0

=> 2x² - 6x + x - 3 = 0

=> 2x(x - 3) + (x - 3) = 0

=> (x-3)(2x+1) = 0

=> x = 3 , - 1/2

Hence,

x = 3 and x = -1/2 are the solutions (roots) of given equation .

Answer 2

$\huge{\boxed{\red{Answer}}}$

$\large{\underline{\pink{Required\;to\;find}}}$

$Whether\;3\;or\;-2\;or\;\dfrac{-1}{2}\;or\;\dfrac{-1}{3}is\;a\;solution\;of\;equation\\2{x}^{2}-5x-3=0$

$\huge{\boxed{\blue{NOTE}}}$

• If 'a' is a root of a given equation then it must satisfy the given equation
• I have used the above statement to verify -3 as root of given equation

$\large{\underline{\purple{Verifying\;3\;as\;a\;solution}}}$

Substituting $3$ in the equation

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

$2({3}^{2})-5(3)-3=0$

$18-15-3=0$

0=0 ( True )

$\boxed{\red{Therefore\;3\;is\;solution\;of\;given\;equation}}$

$\large{\underline{\purple{Verifying\;-2\;as\;a\;solution}}}$

Substituting $-2$ in the equation

$2{x}^{2}-5x-3=0$

$2({-2}^{2})-5(-2)-3=0$

$15=0$

15=0 ( False )

$\boxed{\red{Therefore\;-2\;is'nt\;solution\;of\;given\;equation}}$

$\large{\underline{\purple{Verifying\;\dfrac{-1}{2}\;as\;a\;solution}}}$

Substituting $\dfrac{-1}{2}$ in the equation

$2{x}^{2}-5x-3=0$

$2({(\dfrac{-1}{2})}^{2})-5(\dfrac{-1}{2})-3=0$

$0=0$

0=0 ( True )

$\boxed{\red{Therefore\;\dfrac{-1}{2}\;is\;solution\;of\;given\;equation}}$

$\large{\underline{\purple{Verifying\;\dfrac{-1}{3}\;as\;a\;solution}}}$

Substituting $-2$ in the equation

$2{x}^{2}-5x-3=0$

$2({(\dfrac{-1}{3})}^{2})-5(\dfrac{-1}{3})-3=0$

$\dfrac{-7}{3}=0$

( False )

$\boxed{\red{Therefore\;\dfrac{-1}{3}\;is'nt\;solution\;of\;given\;equation}}$