Hey there! We receieved your request
Stay Tuned as we are going to contact you within 1 Hour
One of our academic counsellors will contact you within 1 working day.
Click to Chat
1800-5470-145
+91 7353221155
Use Coupon: CART20 and get 20% off on all online Study Material
Complete Your Registration (Step 2 of 2 )
Sit and relax as our customer representative will contact you within 1 business day
OTP to be sent to Change
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
= 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
= RHS = x2 - 5x + 6
= (3)2 - 5(3) + 6
= 9 - 15 + 6
= 15 – 15
= 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
= 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
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
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
Similarly putting x = 3 in
LHS ≠ RHS
x = 3 is not the solution the given quadratic equation.
Get your questions answered by the expert for free
You will get reply from our expert in sometime.
We will notify you when Our expert answers your question. To View your Question
Chapter 8: Quadratic Equations Exercise –...