Chapter 8 Exercise 8.1
1. Which of the following are quadratic equations?
(i) x2 +
6x – 4 = 0
Solution:
Let p(x) = x2 + 6x – 4,
It’s clearly seen that p(x) = x2 + 6x – 4 is a
quadratic polynomial. Thus, the given equation is a quadratic equation.
(ii) √3x2 –
2x + 1/2 = 0
Solution:
Let p(x) = √3x2 – 2x + 1/2,
It’s clearly seen that p(x) = √3x2 – 2x + 1/2
having real coefficients is a quadratic polynomial. Thus, the given equation is
a quadratic equation.
(iii) x2 +
1/x2 = 5
Solution:
Given,
x2 + 1/x2 = 5
On multiplying by x2 on both sides we have,
x4 + 1 = 5x2
⇒
x4 –
5x2 +
1 = 0
It’s clearly seen that x4 – 5x2 + 1 is not a
quadratic polynomial as its degree is 4. Thus, the given equation is not a
quadratic equation.
(iv) x – 3/x = x2
Solution:
Given,
x – 3/x = x2
On multiplying by x on both sides we have,
x2 – 3 = x3
⇒
x3 –
x2 +
3 = 0
It’s clearly seen that x3 – x2 + 3 is not a
quadratic polynomial as its degree is 3. Thus, the given equation is not a quadratic
equation.
(v) 2x2 –
√(3x) + 9 = 0
Solution:
It’s clearly seen that 2x2 – √(3x) + 9
is not a polynomial because it contains a term involving x1/2, where 1/2 is not
an integer. Thus, the given equation is not a quadratic equation.
(vi) x2 –
2x – √x – 5 = 0
Solution:
It’s clearly seen that x2 – 2x – √x – 5
is not a polynomial because it contains a term involving x1/2, where 1/2 is not
an integer. Thus, the given equation is not a quadratic equation.
(vii) 3x2 –
5x + 9 = x2 – 7x + 3
Solution:
Given,
3x2 – 5x + 9 = x2 – 7x + 3
On simplifying the equation, we have
2x2 + 2x + 6 = 0
⇒
x2 +
x + 3 = 0 (dividing by 2 on both sides)
Now, it’s clearly seen that x2 + x + 3 is a
quadratic polynomial. Thus, the given equation is a quadratic equation.
(viii) x + 1/x = 1
Solution:
Given,
x + 1/x = 1
On multiplying by x on both sides we have,
x2 + 1 = x
⇒
x2 –
x + 1 = 0
It’s clearly seen that x2 – x + 1 is a
quadratic polynomial. Thus, the given equation is a quadratic equation.
(ix) x2 –
3x = 0
Solution:
Let p(x) = x2 – 3x,
It’s clearly seen that p(x) = x2 – 3x is a
quadratic polynomial. Thus, the given equation is a quadratic equation.
(x) (x + 1/x)2 =
3(x + 1/x) + 4
Solution:
Given,
(x + 1/x)2 = 3(x + 1/x) + 4
⇒
x2 +
1/x2 +
2 = 3x + 3/x + 4
⇒
x4 +
1 + 2x2 =
3x3 +
3x + 4x2
⇒
x4 –
3x3 –
2x2 –
3x + 1 = 0
Now, it’s clearly seen that x4 – 3x3 – 2x2 – 3x + 1 is
not a quadratic polynomial since its degree is 4. Thus, the given equation is
not a quadratic equation.
(xi) (2x + 1)(3x + 2) = 6(x – 1)(x –
2)
Solution:
Given,
(2x + 1)(3x + 2) = 6(x – 1)(x – 2)
⇒
6x2 +
4x + 3x + 2 = 6x2 -12x
– 6x + 12
⇒
7x + 2 = -18x + 12
⇒
25x –
10 = 0
Now, it’s clearly seen that 25x – 10 is not a
quadratic polynomial since its degree is 1. Thus, the given equation is not a
quadratic equation.
(xii) x + 1/x = x2,
x ≠ 0
Solution:
Given,
x + 1/x = x2
On multiplying by x on both sides we have,
x2 + 1 = x3
⇒
x3 –
x2 –
1 = 0
Now, it’s clearly seen that x3 – x2 – 1 is not a
quadratic polynomial since its degree is 3. Thus, the given equation is not a
quadratic equation.
(xiii) 16x2 –
3 = (2x + 5)(5x – 3)
Solution:
Given,
16x2 – 3 = (2x + 5)(5x – 3)
16x2 – 3 = 10x2 – 6x + 25x – 15
⇒
6x2 –
19x + 12 = 0
Now, it’s clearly seen that 6x2 –
19x + 12 is a quadratic polynomial. Thus, the given equation is a quadratic
equation.
(xiv) (x + 2)3 =
x3 – 4
Solution:
Given,
(x + 2)3 = x3 – 4
On expanding, we get
⇒
x3 +
6x2 +
8x + 8 = x3 –
4
⇒
6x2 +
8x + 12 = 0
Now, it’s clearly seen that 6x2 +
8x + 12 is a quadratic polynomial. Thus, the given equation is a quadratic
equation.
(xv) x(x + 1) + 8 = (x + 2)(x – 2)
Solution:
Given,
x(x + 1) + 8 = (x + 2)(x – 2)
x2 + x + 8 = x2 – 4
⇒
x + 12 = 0
Now, it’s clearly seen that x + 12 is a not
quadratic polynomial since its degree is 1. Thus, the given equation is not a
quadratic equation.
2. In each of the following,
determine whether the given values are solutions of the given equation or not:
(i) x2 –
3x + 2 = 0 , x = 2 , x = – 1
Solution:
Here we have,
LHS = x2 – 3x + 2
Substituting x = 2 in LHS, we get
(2)2 – 3(2) + 2
⇒
4 –
6 + 2 = 0 = RHS
⇒
LHS = RHS
Thus, 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
⇒
LHS ≠ RHS
Thus, x = -1 is not a solution of the given equation.
(ii) x2 +
x + 1 = 0, x = 0, x = 1
Solution:
Here we have,
LHS = x2 + x + 1
Substituting x = 0 in LHS, we get
(0)2 + 0 + 1
⇒
1 ≠ RHS
⇒
LHS ≠ RHS
Thus, x = 0 is not a solution of the given
equation.
Similarly,
Substituting x = 1 in LHS, we get
(1)2 + 1 + 1
⇒
3 ≠ RHS
⇒
LHS ≠ RHS
Thus, x = 1 is not a solution of the given
equation.
(iii) x2 −
3√3x + 6 = 0 , x = √3 and x = −2√3
Solution:
Here we have,
LHS = x2 − 3√3x + 6
Substituting x = √3 in LHS, we get
(√3)2 − 3√3(√3) + 6
⇒
3 –
9 + 6 = 0 = RHS
⇒
LHS = RHS
Thus, x = √3 is a solution of the given equation.
Similarly,
Substituting x = −2√3 in LHS, we get
(-2√3)2 − 3√3(-2√3) + 6
⇒
12 + 18 + 6 = 36 ≠
RHS
⇒
LHS ≠
RHS
Thus, x = −2√3 is not a solution of the given
equation.
(iv) x + 1/x = 13/6, x = 5/6, x = 4/3
Solution:
Here we have,
LHS = x +1/ x
Substituting x = 5/6 in LHS, we get
(5/6) + 1/(5/6) = 5/6 + 6/5
⇒
61/30 ≠ RHS
⇒
LHS ≠ RHS
Thus, x = 5/6 is not a solution of the given
equation.
Similarly,
Substituting x = 4/3 in LHS, we get
(4/3) + 1/(4/3) = 4/3 + 3/4
⇒
25/12 ≠ RHS
⇒
LHS ≠ RHS
Thus, x = 4/3 is not a solution of the given
equation.
(v) 2x2 –
x + 9 = x2 + 4x + 3, x = 2, x = 3
Solution:
Here we have,
2x2 – x + 9 = x2 + 4x + 3
⇒
x2 –
5x + 6 = 0
LHS = x2 – 5x + 6
Substituting x = 2 in LHS, we get
(2)2 – 5(2) + 6
⇒
4 –
10 + 6 = 0 = RHS
⇒
LHS = RHS
Thus, x = 2 is a solution of the given equation.
Similarly,
Substituting x = 3 in LHS, we get
(3)2 – 5(3) + 6
⇒
9 –
15 + 6 = 0 = RHS
⇒
LHS = RHS
Thus, x = 3 is a solution of the given equation.
(vi) x2 –
√2x – 4 = 0, x = -√2, x = -2√2
Solution:
Here we have,
LHS = x2 – √2x – 4
Substituting x = -√2 in LHS, we get
(-√2)2 − √2(-√2) – 4
⇒
4 + 2 –
4 = 2 ≠
RHS
⇒
LHS ≠
RHS
Thus, x = -√2 is a solution of the given equation.
Similarly,
Substituting x = −2√2 in LHS, we get
(-2√2)2 − √2(-2√2) – 4
⇒
8 + 4 –
4 = 8 ≠
RHS
⇒
LHS ≠
RHS
Thus, x = −2√2 is not a solution of the given
equation.
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