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Class 10 RD Sharma Solutions – Chapter 8 Quadratic Equations – Exercise 8.1
• Last Updated : 28 Dec, 2020

### (i) x2 + 6x – 4 = 0

Solution:

As we know that, quadratic equation is in form of: ax2 + bx + c = 0

Here, given equation is x2 + 6x – 4 =0  in quadratic equation.

So, It is a quadratic equations.

### (ii) √3x2 − 2x + ½= 0

Solution:

As we know that, quadratic equation is in form of: ax2 + bx + c = 0

Here, given equation is √3x2 − 2x + ½= 0 in quadratic equation.

So, It is a quadratic equation.

### (iii) x2 + 1/x2  = 5

Solution:

This equation can be written as: x4 – 5x2 + 1 =0

As we know that, quadratic equation is in form of: ax2 + bx + c = 0

Here, given equation is x4 – 5x2 + 1 =0 is 4 degree polynomial.

So, It is not quadratic equations.

### (iv) x – 3/x = x2

Solution:

This equation can be written as: x3 – x2 – 3 = 0

As we know that, quadratic equation is in form of: ax2 + bx + c = 0

Here, given equation is x3 – x2 – 3 = 0 is 3 degree polynomial.

So, It is not quadratic equations.

### (v) 2x2 – √3x + 9 = 0

Solution:

As we know that, quadratic equation is in form of: ax2 + bx + c = 0

Here, given equation is 2x2 – √3x + 9 = 0  in quadratic equation.

So, It is a quadratic equation.

### (vi) x2 – 2x -√x – 5 = 0

Solution:

As we know that, quadratic equation is in form of: ax2 + bx + c = 0

Here, given equation is x2 – 2x -√x -5 = 0.

So, It is not quadratic equation.

### (vii) 3x2 − 5x + 9 = x2 − 7x + 3

Solution:

This equation can be written as: 2x2 + 2x + 6 = 0

As we know that, quadratic equation is in form of: ax2 + bx + c = 0

Here, given equation is 2x2 + 2x + 6 = 0 in quadratic equation.

So, It is a quadratic equation.

### (viii) x + 1/x = 1

Solution:

This equation can be written as: x2 – x + 1 = 0

As we know that, quadratic equation is in form of: ax2 + bx + c = 0

Here, given equation is x2 – x + 1 = 0 in quadratic equation.

So, It is a quadratic equation.

### (ix) x2 – 3x = 0

Solution:

As we know that, quadratic equation is in form of: ax2 + bx + c = 0

Here, given equation is  x2 – 3x =0  in quadratic equation.

So, It is a quadratic equation.

### (x) (x + 1/x)2 = 3(x +1/x)

Solution:

((x2 + 1)/x)2 = 3((x2 + 1)/2)

This equation can be written as: x5 – 3x4 + 2x3 – 3x2 + x =0

As we know that, quadratic equation is in form of: ax2 + bx + c = 0

Here, given equation is x5 – 3x4 + 2x3 – 3x2 + x = 0 is 5 degree polynomial.

So, It is not quadratic equations.

### (xi) (2x2 + 1)(3x2 + 2) = 6(x2− 1)(x2− 2)

Solution:

This equation can be written as:

6x2 + 4x + 3x + 2 = 6x2 -12x – 6x + 12

7x + 2 = -18x + 12

This equation can be written as: 25x – 10 = 0

As we know that , quadratic equation is in form of: ax2 + bx + c = 0

Here, given equation is 25x – 10 = 0 is 1 degree polynomial.

So, It is not quadratic equations.

### (xii) x + 1/x = x2, x ≠ 0

Solution:

x + 1/x = x2

On multiplying by x on both sides we have,
x2 + 1 = x3

This equation can be written as: x3 – x2 – 1 = 0

As we know that, quadratic equation is in form of: ax2 + bx + c = 0

Here, given equation x3 – x2 – 1 = 0 is  = 0 is 3 degree polynomial.

So, It is not quadratic equations.

### (xiii) 16x2 – 3 = (2x + 5)(5x – 3)

Solution:

16x2 – 3 = (2x + 5)(5x – 3)

16x2 – 3 = 10x2 – 6x + 25x – 15

This equation can be written as: 6x2 – 19x + 12 = 0

As we know that , quadratic equation is in form of: ax2 + bx + c = 0

Here, given equation is 6x2 – 19x + 12 = 0 is 2 degree polynomial.

So, It is quadratic equations.

### (xiv) (x + 2)3 = x3 – 4

Solution:

(x + 2)3 = x3 – 4

On expanding, we get
x3 + 6x2 + 8x + 8 = x3 – 4

6x2 + 8x + 12 = 0

This equation can be written as: 6x2 + 8x + 12 = 0

As we know that, quadratic equation is in form of: ax2 + bx + c = 0

Here, given equation is 6x2 – 19x + 12 = 0 is 2 degree polynomial.

So, It is quadratic equations.

### (xv) x(x + 1) + 8 = (x + 2)(x – 2)

Solution:

x(x + 1) + 8 = (x + 2)(x – 2)

x2 + x + 8 = x2 – 4

x + 12 = 0

This equation can be written as: x + 12 = 0

As we know that, quadratic equation is in form of: ax2 + bx + c = 0

Here, given equation is x + 12 = 0 is 0 degree polynomial.

So, It is not a quadratic equation.

### (i) x2 – 3x + 2 = 0, x = 2, x = – 1

Solution:

Here we have check values to determine the solution the given equation:

LHS = x2– 3x + 2

Now we will put x = 2 in LHS, we get

(2)2 – 3(2) + 2

⇒ 4 – 6 + 2 = 0 = RHS

⇒ LHS = RHS

So, we can say that x = 2 is not a solution of the given equation.

Similarly,

Now, we will put x = -1 in LHS, we get

(-1)2 – 3(-1) + 2

1 + 3 + 2 = 6 ≠ RHS

LHS ≠ RHS

So, we can say that x = -1 is not a solution of the given equation.

### (ii) x2 + x + 1 = 0, x = 0, x = 1

Solution:

Here we have check values to determine the solution the given equation:

LHS = x2 + x + 1

Now we will put x = 0 in LHS, we get

(0)2 + 0 + 1

⇒ 1 ≠ RHS

⇒ LHS ≠ RHS

So , we can say that x = 0 is not a solution of the given equation.

Similarly,

Now, we will put x = 1 in LHS, we get

(1)2 + 1 + 1

⇒ 3 ≠ RHS

⇒ LHS ≠ RHS

So, we can say that 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 check values to determine the solution the given equation:

LHS = x2 − 3√3x + 6

Now, we will put x = √3 in LHS, we get

(√3)2 − 3√3(√3) + 6

⇒ 3 – 9 + 6 = 0 = RHS

⇒ LHS = RHS

So , we can say that x = √3 is a solution of the given equation.

Similarly,

Now, we will put x = −2√3 in LHS, we get

(-2√3)2 − 3√3(-2√3) + 6

⇒ 12 + 18 + 6 = 36 ≠ RHS

⇒ LHS ≠ RHS

So, we can say that 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 check values to determine the solution the given equation:

LHS = x +1/ x

Now, we will put x = 5/6 in LHS, we get

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

⇒ 61/30 ≠ RHS

⇒ LHS ≠ RHS

So , we can say that x = 5/6 is not a solution of the given equation.

Similarly,

Now, we will put x = 4/3 in LHS, we get

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

⇒ 25/12 ≠ RHS

⇒ LHS ≠ RHS

So, we can say that x = 3/4 is not a solution of the given equation.

### (v) 2x2 – x + 9 = x2 + 4x + 3, x = 2, x = 3

Solution:

Here we have check values to determine the solution the given equation:

2x2– x + 9 = x2 + 4x + 3

⇒ x2 – 5x + 6 = 0

LHS = x2– 5x + 6

Now, we will put x = 2 in LHS, we get

(2)2 – 5(2) + 6

4 – 10 + 6 = 0 = RHS

⇒ LHS = RHS

So , we can say that x = 2 is a solution of the given equation.

Similarly,

Now, we will put x = 3 in LHS, we get

(3)2 – 5(3) + 6

⇒ 9 – 15 + 6 = 0 = RHS

⇒ LHS = RHS

So, we can say that x = 3 is a solution of the given equation.

### (vi) x2 – √2x – 4 = 0, x = -√2, x = -2√2

Solution:

Here we have check values to determine the solution the given equation:

LHS = x2 – √2x – 4

Now, we will put x = -√2 in LHS, we get

(-√2)2 − √2(-√2) – 4

⇒ 4 + 2 – 4 = 2 ≠ RHS

⇒ LHS ≠ RHS

So , we can say that x = -√2 is a solution of the given equation.

Similarly,

Now, we will put x = −2√2 in LHS, we get

(-2√2)2 − √2(-2√2) – 4

⇒ 8 + 4 – 4 = 8 ≠ RHS

⇒ LHS ≠ RHS

So, we can say that x = −2√2 is not a solution of the given equation.

### (vii) a2x2 – 3abx + 2b2 = 0, x = a/b, x = b/a

Solution:

Here we have check values to determine the solution the given equation:

LHS =  a2x2 – 3abx + 2b2

Now, we will put x = a/b in LHS, we get

a2(a/b)2 -3ab(a/b) + 2b2

⇒ a4/b2 – 3a2 + 2b2 ≠ RHS

⇒ LHS ≠ RHS

So, we can say that x = a/b is a solution of the given equation.

Similarly,

Now, we will put x = b/a in LHS, we get

a2(b/a)2 – 3ab(b/a) + 2b2

⇒ b2 – 3b2 + 2b2 = RHS

⇒ LHS = RHS

So, we can say that x = b/a is not a solution of the given equation.

### (i) 7x2 + kx – 3 = 0, x = 2/3

Solution:

Here we have to find value of k. So, we will put value of x = 2/3 in the given equation.

Here, x = 2/3

7(2/3)2 + k(2/3) – 3 =0

28/9 + 2k/3 -3 =0

2k/3 = 3 – 28/9

2k/3 = (27 – 28)/9

2k/3 = -1/9

k = -1/6

On putting value of x = 2/3 , we will get k =-1/6.

### (ii) x2 – x(a + b) + k = 0, x = a

Solution:

Here we have to find value of k. So, we will put value of x = a in the given equation.

Here, x = a

a2 – a2 – ab + k = 0

k = ab

On putting value of x = a, we get k = ab.

### (iii) kx2 + √2x – 4 = 0, x = √2

Solution:

Here we have to find value of k. So, we will put value of x = √2 in the given equation.

Here, x = √2

k(√2)2 + (√2)2 – 4 = 0

2k + 2 – 4 = 0

2k – 2 = 0

k = 1

On putting value of x = √2 , we get k = 1.

### (iv) x2 + 3ax + k = 0, x = -a

Solution:

Here we have to find value of k. So, we will put value of x = -a in the given equation.

Here, x = -a

(- a)2 + 3a(- a) + k = 0

a2 – 3a2 + k = 0

k = 2a2

On putting value of x = – a, we get k = 2a2.

### √(x2 – 4x +3) + √(x2 -9) = √(4x2 – 14x + 16)

Solution:

Here we have to check, 3 is root of equation or not.

So, we first put x = 3 in LHS side.

√((3)2 – 4(3) + 3) + √((3)2 – 9)

⇒ 0 + 0 = 0

Now , we will put value x = 3 in RHS side.

√(4(3)2 – 14(3) + 16)

⇒ √(52 – 42) = √10

Now we can say that, LHS ≠ RHS.

So, x = 3 is not the root of the equation.

### Question 5. If x = 2/3 and x = -3 are the roots of the equation ax2 + 7x + b = 0, find the values of a and b.

Solution:

Here we have to find value of a and b, so we will put values x = 2/3 and x = -3 in the equation.

So first put x = 2/3

a(2/3)2 + 7(2/3) + b =0

4a/9 + 14/3 + b =0

4a + 42 + 3b = 0     — (i)

Now, x = -3.

a(-3)2 + 7(-3) + b = 0

9a – 21 + b = 0       –(ii)

Now solving these equations (i) & (ii).

After solving these equations, we get a = 3 and b = -6.

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