# Class 10 RD Sharma Solutions – Chapter 8 Quadratic Equations – Exercise 8.1

### (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) âˆš3x2 âˆ’ 2x + Â½= 0

Solution:

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

Here, given equation is âˆš3x2 âˆ’ 2x + Â½= 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 – âˆš3x + 9 = 0

Solution:

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

Here, given equation is 2x2 – âˆš3x + 9 = 0  in quadratic equation.

So, It is a quadratic equation.

### (vi) x2 – 2x -âˆšx – 5 = 0

Solution:

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

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

So, It is not quadratic equation.

### (vii) 3x2 âˆ’ 5x + 9 = x2 âˆ’ 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(x2âˆ’ 1)(x2âˆ’ 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 â€“ 10 = 0

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

Here, given equation is 25x â€“ 10 = 0 is 1 degree polynomial.

So, It is not quadratic equations.

### (xii) x + 1/x = x2, x â‰  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 â€“ x2 â€“ 1 = 0

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

Here, given equation x3 â€“ x2 â€“ 1 = 0 is  = 0 is 3 degree polynomial.

So, It is not quadratic equations.

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

Solution:

16x2 â€“ 3 = (2x + 5)(5x – 3)

16x2 â€“ 3 = 10x2 â€“ 6x + 25x â€“ 15

This equation can be written as: 6x2 â€“ 19x + 12 = 0

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

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

### (xiv) (x + 2)3 = x3 â€“ 4

Solution:

(x + 2)3 = x3 â€“ 4

On expanding, we get x3 + 6x2 + 8x + 8 = x3 â€“ 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 â€“ 19x + 12 = 0 is 2 degree polynomial.

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

Solution:

x(x + 1) + 8 = (x + 2)(x – 2)

x2 + x + 8 = x2 â€“ 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 â€“ 3(2) + 2

â‡’ 4 â€“ 6 + 2 = 0 = RHS

â‡’ 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 â€“ 3(-1) + 2

1 + 3 + 2 = 6 â‰  RHS

LHS â‰  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

â‡’ 1 â‰  RHS

â‡’ LHS â‰  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

â‡’ 3 â‰  RHS

â‡’ LHS â‰  RHS

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

### (iii) x2 âˆ’ 3âˆš3x + 6 = 0, x = âˆš3 and x = âˆ’2âˆš3

Solution:

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

LHS = x2 âˆ’ 3âˆš3x + 6

Now, we will put x = âˆš3 in LHS, we get

(âˆš3)2 âˆ’ 3âˆš3(âˆš3) + 6

â‡’ 3 â€“ 9 + 6 = 0 = RHS

â‡’ LHS = RHS

So , we can say that x = âˆš3 is a solution of the given equation.

Similarly,

Now, we will put x = âˆ’2âˆš3 in LHS, we get

(-2âˆš3)2 âˆ’ 3âˆš3(-2âˆš3) + 6

â‡’ 12 + 18 + 6 = 36 â‰  RHS

â‡’ LHS â‰  RHS

So, we can say that x = âˆ’2âˆš3 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

â‡’ 61/30 â‰  RHS

â‡’ LHS â‰  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

â‡’ 25/12 â‰  RHS

â‡’ LHS â‰  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

â‡’ x2 â€“ 5x + 6 = 0

LHS = x2– 5x + 6

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

(2)2 â€“ 5(2) + 6

4 â€“ 10 + 6 = 0 = RHS

â‡’ 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 â€“ 5(3) + 6

â‡’ 9 â€“ 15 + 6 = 0 = RHS

â‡’ LHS = RHS

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

### (vi) x2 â€“ âˆš2x â€“ 4 = 0, x = -âˆš2, x = -2âˆš2

Solution:

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

LHS = x2 â€“ âˆš2x â€“ 4

Now, we will put x = -âˆš2 in LHS, we get

(-âˆš2)2 âˆ’ âˆš2(-âˆš2) â€“ 4

â‡’ 4 + 2 â€“ 4 = 2 â‰  RHS

â‡’ LHS â‰  RHS

So , we can say that x = -âˆš2 is a solution of the given equation.

Similarly,

Now, we will put x = âˆ’2âˆš2 in LHS, we get

(-2âˆš2)2 âˆ’ âˆš2(-2âˆš2) – 4

â‡’ 8 + 4 – 4 = 8 â‰  RHS

â‡’ LHS â‰  RHS

So, we can say that x = âˆ’2âˆš2 is not a solution of the given equation.

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

Solution:

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

LHS =  a2x2 â€“ 3abx + 2b2

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

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

â‡’ a4/b2 – 3a2 + 2b2 â‰  RHS

â‡’ LHS â‰  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

â‡’ b2 – 3b2 + 2b2 = RHS

â‡’ 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 + âˆš2x – 4 = 0, x = âˆš2

Solution:

Here we have to find value of k. So, we will put value of x = âˆš2 in the given equation.

Here, x = âˆš2

k(âˆš2)2 + (âˆš2)2 – 4 = 0

2k + 2 â€“ 4 = 0

2k â€“ 2 = 0

k = 1

On putting value of x = âˆš2 , 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.

### âˆš(x2 – 4x +3) + âˆš(x2 -9) = âˆš(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.

âˆš((3)2 – 4(3) + 3) + âˆš((3)2 – 9)

â‡’ 0 + 0 = 0

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

âˆš(4(3)2 – 14(3) + 16)

â‡’ âˆš(52 – 42) = âˆš10

Now we can say that, LHS â‰  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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