Question

In each of the following, determine whether the given values are solutions of the given equation or not:
(iv)$x+\frac{1}{x}= \frac{13}{6}$, $x=\frac{5}{6}$,$x=\frac{4}{3}$
(v)2x² -x+9=x² +4x+3, x=2, x=3
(vi)x² -2x-4=0, x=-√2, x=-2√2
(vii)a² x² -3abx+2b² =0, $x=\frac{a}{b}$, $x=\frac{b}{a}$

Answer 1

SOLUTION  :

SOLUTION of part (iv) & (vii) are in the attachment.

(v) Given :  2x²  - x + 9 = x² + 4x + 3

= 2x² - x² - x - 4x + 9 - 3  

= x² - 5x + 6 = 0

Equation is of the form p(x) = 0 , where  

p(x) = x² - 5x + 6 …………….(1)

On putting x = 2 in eq 1,

p(2) = (2)² - 5(2) + 6

= 4 - 10 + 6

= - 6 + 6

= 0

p(2) = 0

So, x = 2 is the solution of the given equation.

Now on putting x = 3  in eq 1,

p(3) = x² - 5x + 6

= (3)² - 5(3) + 6

= 9 - 15 + 6

= 6 – 6

= 0

p(3) = 0

So, x =3 is  the solution of the given equation.

Hence, x = 2 and  x = 3  both are the solution of the above equation  

(vi) Given : x² - √2x - 4 = 0

Equation is of the form p(x) = 0 , where  

p(x) = x² - √2x - 4  …………….(1)

On putting x = - √2 in eq 1,

p(- √2) = (-√2)² - √2 × -√2 - 4

= 2 + 2 - 4  

= 4 - 4  

= 0

p(-√2) = 0

So, x = -√2 is  the solution of the given equation.

Now on putting x = -2√2  in eq 1,

p(x) = x² - √2x - 4

p(- 2√2) = (-2√2)² - √2 × -2√2 - 4

= 8 + 4 - 4

= 12 - 4  

= 8  

p(-2√2) ≠ 0

So, x = -2√2 is not the solution of the given equation.

Hence, x = - √2 is a solution of the above equation and  x = - 2√2 is not a solution of the above  equation.  

HOPE THIS ANSWER WILL HELP YOU…

Answer 2

_______________________

We know that ,

A solution to an equation is a value that, when substituted for the variable, makes the equation true.

__________________________

(iv)$x+\frac{1}{x}= \frac{13}{6}$, $x=\frac{5}{6}$,$x=\frac{4}{3}$

a)Substitute x = 5/6 in given

equation ,

LHS =(5/6 )+ 1/(5/6)

= 5/6 + 6/5

= ( 30 + 36 )/30

= 66/30

= 11/5

≠ RHS

Therefore ,

x = 5/6 is not a solution.

b) substitute x = 4/3

LHS = 4/3 + 1/(4/3)

= 4/3 + 3/4

= ( 16 + 9 )/12

= 25/12

≠ RHS

(v)2x² -x+9=x² +4x+3, x=2, x=3

a) substitute x = 2 in the

equation ,

LHS = 2(2)²-2+9

= 8 - 2 + 9

= 15

RHS = 2²+4×2+3

= 4 + 8 + 3

= 15

LHS = RHS

Therefore ,

x = 2 is a solution .

b) Substitute x = 3 in the

equation,

LHS = 2(3)²-3+9

= 18 + 6

= 24

RHS = 3²+4×3+3

= 9 + 12 + 3

= 24

LHS = RHS

Therefore ,

x = 3 is a solution.

(vi)x² -2x-4=0, x=-√2, x=-2√2

a) Substitute x = √2 , in the

equation,

LHS = (√2)²-2(√2)-4

= 4 - 2√2 - 4

= -2√2

≠ 0

≠ RHS

x = √2 is not a solution.

b) substitute x = -2√2,

LHS = (-2√2)²-2(-2√2)-4

= 8 + 4√2 - 4

= 4 + 4√2

≠0

≠ RHS

Therefore ,

x = -2√2 is not a solution.

vii)a² x² -3abx+2b² =0, $x=\frac{a}{b}$, $x=\frac{b}{a}$

a) substitute x = a/b

in the equation,

LHS = a²(a/b)²-3ab(a/b)+2b²

= a⁴/b² - 3a² + 2b²

≠ 0

≠ RHS

Therefore ,

x = a/b is not a solution.

b) substitute x = b/a in the

equation,

LHS = a²(b/a)²-3ab(b/a)+2b²

= b² - 3b² + 2b²

= 0

= RHS

Therefore ,

x = b/a is a solution.

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