Solve quadratic equations:
(i) x2 - 3x + 2 = 0 , x = 2 , x = - 1
(ii) x2 + x + 1 = 0, x = 0, x = 1
(v) 2x2 - x + 9 = x2 + 4x + 3, x = 2 and x = 3
(vii) a2x2 - 3abx + 2b2 = 0
(i) Here LHS = x2 - 3x + 2 RHS = 0
Now, substitute x = 2 in LHS
We get, (2)2 - 3(2) + 2
= 4 - 6 + 2
= 6 - 6
= 0
RHS Since, LHS =RHS
Therefore, x - 2 is a solution of the given equation.
Similarly, substituting x = - 1 in
LHS
We get, (-1)2 - 3(-1) + 2
= 1 + 3 + 2
= 6
RHS Since, LHS?
RHS = x = - 1 is not the solution of the given equation.
(ii) Here,
LHS = x2 + x + 1 and RHS = 0
Now, substituting x = 0 and x = 1 in
LHS = 02 + 0 + 1
= (1)2 + 1 + 1
= 1 + 2
= 3 LHS
RHS Both x = 0 and x = 1 are not solutions of the given equation.
(iii) Here,
LHS = x2 - 3√3x + 6 = 0 and
RHS = 0 Substituting the value of x = √3 and x = - 2√3 in
= 3 – 9 + 6
= 9 - 9
= 0
= 12 + 18 + 6
= 36
RHS x = √3 is a solution of the above mentioned equation. Whereas, x = - 2√3 is not a solution of the above mentioned equation.
(iv) Here,
Substituting where x = 5/6 and x = 4/3 in the LHS
RHS where x = 5/6 and x = 4/3 are not the solutions of the given equation.
(v) 2x2 - x + 9 = x2 + 4x + 3, x = 2
= 2x2 - x + 9 - x2 + 4x + 3 = x2 - 5x + 6 = 0
Here, LHS = x2 - 5x + 6 and RHS = 0
Substituting x = 2 and x = 3
= x2 - 5x + 6
= (2)2 - 5(2) + 6
= 10 - 10
= 0
= RHS = x2 - 5x + 6
= (3)2 - 5(3) + 6
= 9 - 15 + 6
= 15 – 15
= 0
= RHS x = 2 and x = 3 both are the solutions of the given quadratic equation.
(vi) Here,
Substituting the value x = - √2 and x = - 2√2 in
= 2 – 2 – 4
= – 4
= 8 – 4 – 4
= 8 – 8
= 0
= RHS x = - 2√2 is the solution of the above mentioned quadratic equation.
(vii) Here,
LHS = a2x2 - 3abx + 2b2 and RHS = 0
RHS = x = b/a
is the solution of the above mentioned quadratic equation.
(i) Given that 2/3 is a root of the given equation. The equation is 7x2 + kx - 3 = 0
(ii) Given that x=a is a root of the given equation x2 - x(a + b) + k = 0
(iii) Given that x = √2 is a root of the given equation kx2 + √2x - 4
(iv) Given that x = - a is the root of the given equation x2 + 3ax + k = 0
(i) Given that 2/3 is a root of the given equation. The equation is 7x2 + kx - 3 = 0
According to the question 2/3 satisfies the equation.
(ii) Given that x = a is a root of the given equation x2 - x(a + b) + k = 0 = x = a satisfies the equation = a2- a(a + b) + k = 0
= a2 - a2- ab + k
K = ab
(iii) Given that x = √2 is a root of the given equation kx2 + √2x - 4
x = √2 satisfies the given quadratic equation.
= 2k + 2 – 4 = 0
= 2k – 2 = 0
k = 1
(iv) Given that x= -a is the root of the given equation x2 + 3ax + k = 0
Therefore, = (- a)2 + 3a(- a) + k = 0
= a2 + 3a2 + k = 0
= k = 4a2= - a satisfies the equation
Given to check whether 3 is a root of the equation
= 0
Similarly putting x = 3 in
LHS ≠ RHS
x = 3 is not the solution the given quadratic equation.