Degree 2 polynomials when equated to zero are called quadratic equations. These quadratic equations have certain conditions to satisfy in order to exist. Students build up strong basic knowledge by referring to RD Sharma Solutions Class 10, which are prepared by subject experts at BYJU’S. Also, the RD Sharma Solutions for Class 10 Maths Chapter 8 Quadratic Equations Exercise 8.1 PDF provided below is available for detailed references.

## RD Sharma Solutions for Class 10 Chapter 8 Quadratic Equations Exercise 8.1 Download PDF

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

**1. Which of the following are quadratic equations?**

**(i) x ^{2} + 6x â€“ 4 = 0 **

**Solution: **

Let p(x) = x^{2} + 6x â€“ 4,

Itâ€™s clearly seen that p(x) = x^{2} + 6x â€“ 4 is a quadratic polynomial. Thus, the given equation is a quadratic equation.

**(ii) âˆš3x ^{2} â€“ 2x + 1/2 = 0**

**Solution: **

Let p(x) = âˆš3x^{2} â€“ 2x + 1/2,

Itâ€™s clearly seen that p(x) = âˆš3x^{2} â€“ 2x + 1/2 having real coefficients is a quadratic polynomial. Thus, the given equation is a quadratic equation.

**(iii) x ^{2} + 1/x^{2} = 5 **

**Solution: **

** **

Given,

x^{2} + 1/x^{2} = 5

On multiplying by x^{2 }on both sides we have,

x^{4} + 1 = 5x^{2}

â‡’ x^{4} â€“ 5x^{2} + 1 = 0

Itâ€™s clearly seen that x^{4} â€“ 5x^{2} + 1 is not a quadratic polynomial as its degree is 4. Thus, the given equation is not a quadratic equation.

**(iv) x â€“ 3/x = x ^{2}**

**Solution: **

Given,

x â€“ 3/x = x^{2}

On multiplying by x^{ }on both sides we have,

x^{2} â€“ 3 = x^{3}

â‡’ x^{3} â€“ x^{2} + 3 = 0

Itâ€™s clearly seen that x^{3} â€“ x^{2} + 3 is not a quadratic polynomial as its degree is 3. Thus, the given equation is not a quadratic equation.

**(v) 2x ^{2} – âˆš(3x) + 9 = 0**

**Solution: **

Itâ€™s clearly seen that 2x^{2} – âˆš(3x) + 9 is not a polynomial because it contains a term involving x^{1/2}, where 1/2 is not an integer. Thus, the given equation is not a quadratic equation.

**(vi) x ^{2} â€“ 2x – âˆšx â€“ 5 = 0**

**Solution: **

** **

Itâ€™s clearly seen that x^{2} â€“ 2x – âˆšx â€“ 5** **is not a polynomial because it contains a term involving x^{1/2}, where 1/2 is not an integer. Thus, the given equation is not a quadratic equation.

**(vii) 3x ^{2} â€“ 5x + 9 = x^{2} â€“ 7x + 3**

**Solution: **

** **

Given,

3x^{2} â€“ 5x + 9 = x^{2} â€“ 7x + 3

On simplifying the equation, we have

2x^{2} + 2x + 6 = 0

â‡’ x^{2} + x + 3 = 0 (dividing by 2 on both sides)

Now, itâ€™s clearly seen that x^{2} + 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,

x^{2} + 1 = x

â‡’ x^{2} â€“ x + 1 = 0

Itâ€™s clearly seen that x^{2} â€“ x + 1 is a quadratic polynomial. Thus, the given equation is a quadratic equation.

**(ix) x ^{2} â€“ 3x = 0 **

**Solution: **

Let p(x) = x^{2} â€“ 3x,

Itâ€™s clearly seen that p(x) = x^{2} â€“ 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

â‡’ x^{2} + 1/x^{2} + 2 = 3x + 3/x + 4

â‡’ x^{4} + 1 + 2x^{2} = 3x^{3} + 3x + 4x^{2}

â‡’ x^{4} – 3x^{3} â€“ 2x^{2} â€“ 3x + 1 = 0

Now, itâ€™s clearly seen that x^{4} â€“ 3x^{3} â€“ 2x^{2} â€“ 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 â€“ 1)(x â€“ 2)**

**Solution: **

Given,

(2x + 1)(3x + 2) = 6(x â€“ 1)(x â€“ 2)

â‡’ 6x^{2} + 4x + 3x + 2 = 6x^{2} -12x – 6x + 12

â‡’ 7x + 2 = -18x + 12

â‡’ 25x â€“ 10 = 0

Now, itâ€™s clearly seen that 25x â€“ 10 is not a quadratic polynomial since its degree is 1. Thus, the given equation is not a quadratic equation.

** **

**(xii) x + 1/x = x ^{2}, x â‰ 0**

**Solution: **

** **

** **Given,

x + 1/x = x^{2}

On multiplying by x^{ }on both sides we have,

x^{2} + 1 = x^{3}

â‡’ x^{3} â€“ x^{2} â€“ 1 = 0

Now, itâ€™s clearly seen that x^{3} â€“ x^{2} â€“ 1 is not a quadratic polynomial since its degree is 3. Thus, the given equation is not a quadratic equation.

**(xiii) 16x ^{2} â€“ 3 = (2x + 5)(5x – 3)**

**Solution: **

** **

** **Given,

16x^{2} â€“ 3 = (2x + 5)(5x – 3)

16x^{2} â€“ 3 = 10x^{2} â€“ 6x + 25x â€“ 15

â‡’ 6x^{2 }â€“ 19x + 12 = 0

Now, itâ€™s clearly seen that 6x^{2 }â€“ 19x + 12 is a quadratic polynomial. Thus, the given equation is a quadratic equation.

**(xiv) (x + 2) ^{3} = x^{3} â€“ 4 **

**Solution: **

** **

** **Given,

(x + 2)^{3} = x^{3} â€“ 4

On expanding, we get

â‡’ x^{3} + 6x^{2} + 8x + 8 = x^{3} â€“ 4

â‡’ 6x^{2 }+ 8x + 12 = 0

Now, itâ€™s clearly seen that 6x^{2 }+ 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)**

x^{2} + x + 8 = x^{2} â€“ 4

â‡’ x + 12 = 0

Now, itâ€™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) x ^{2Â Â }– 3x + 2 = 0 , x = 2 , x = – 1**

**Solution: **

** **

Here we have,

LHS = x^{2Â }– 3x + 2

Substituting x = 2 in LHS, we get

(2)^{2Â }â€“ 3(2) + 2

â‡’ 4 â€“ 6 + 2 = 0 = RHS

â‡’ LHS = RHS

Thus, x = 2 is a solution of the given equation.

Similarly,

Substituting x = -1 in LHS, we get

(-1)^{2Â }â€“ 3(-1) + 2

â‡’ 1 + 3 + 2 = 6 **â‰ **RHS

â‡’ LHS **â‰ **RHS

Thus, x = -1 is not a solution of the given equation.

**(ii) Â x ^{2Â }+ x + 1 = 0, x = 0, x = 1**

**Solution: **

** **

Here we have,

LHS = x^{2Â }+ x + 1

Substituting x = 0 in LHS, we get

(0)^{2Â }+ 0 + 1

â‡’ 1 **â‰ ** RHS

â‡’ LHS **â‰ ** RHS

Thus, x = 0 is not a solution of the given equation.

Similarly,

Substituting x = 1 in LHS, we get

(1)^{2Â }+ 1 + 1

â‡’ 3 **â‰ **RHS

â‡’ LHS **â‰ **RHS

Thus, x = 1 is not a solution of the given equation.

**(iii) x ^{2 }âˆ’ 3âˆš3x + 6 = 0Â ,Â x = âˆš3 and x = âˆ’2âˆš3**

**Solution: **

Here we have,

LHS = x^{2 }âˆ’ 3âˆš3x + 6

Substituting x = âˆš3 in LHS, we get

(âˆš3)^{2 }âˆ’ 3âˆš3(âˆš3) + 6

â‡’ 3 â€“ 9 + 6 = 0 = RHS

â‡’ LHS = RHS

Thus, x = âˆš3 is a solution of the given equation.

Similarly,

Substituting x = âˆ’2âˆš3 in LHS, we get

(-2âˆš3)^{2 }âˆ’ 3âˆš3(-2âˆš3) + 6

â‡’ 12 + 18 + 6 = 36 â‰ RHS

â‡’ LHS â‰ RHS

Thus, 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,

LHS = x^{Â }+1/ x

Substituting x = 5/6 in LHS, we get

(5/6)^{Â }+ 1/(5/6) = 5/6 + 6/5

â‡’ 61/30 **â‰ ** RHS

â‡’ LHS **â‰ ** RHS

Thus, x = 5/6 is not a solution of the given equation.

Similarly,

Substituting x = 4/3 in LHS, we get

(4/3)^{Â }+ 1/(4/3) = 4/3 + 3/4

â‡’ 25/12 **â‰ **RHS

â‡’ LHS **â‰ **RHS

Thus, x = 3/4 is not a solution of the given equation.

**(v) Â 2x ^{2Â }– x + 9 = x^{2Â }+ 4x + 3, x = 2, x = 3**

**Solution: **

** **

Here we have,

2x^{2Â }– x + 9 = x^{2Â }+ 4x + 3

â‡’ x^{2 }â€“ 5x + 6 = 0

LHS = x^{2Â }– 5x + 6

Substituting x = 2 in LHS, we get

(2)^{2Â }â€“ 5(2) + 6

â‡’ 4 â€“ 10 + 6 = 0 **=** RHS

â‡’ LHS **=** RHS

Thus, x = 2 is a solution of the given equation.

Similarly,

Substituting x = 3 in LHS, we get

(3)^{2Â }â€“ 5(3) + 6

â‡’ 9 â€“ 15 + 6 = 0 **= **RHS

â‡’ LHS **= **RHS

Thus, x = 3 is a solution of the given equation.

**(vi) x ^{2} â€“ âˆš2x â€“ 4 = 0, x = -âˆš2, x = -2âˆš2 **

**Solution: **

Here we have,

LHS = x^{2} â€“ âˆš2x â€“ 4

Substituting 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

Thus, x = -âˆš2 is a solution of the given equation.

Similarly,

Substituting x = âˆ’2âˆš2 in LHS, we get

(-2âˆš2)^{2 }âˆ’ âˆš2(-2âˆš2) – 4

â‡’ 8 + 4 – 4 = 8 â‰ RHS

â‡’ LHS â‰ RHS

Thus, x = âˆ’2âˆš2 is not a solution of the given equation.

**(vii) a ^{2}x^{2} â€“ 3abx + 2b^{2} = 0, x = a/b, x = b/a **

**Solution: **

We have,

LHS = a^{2}x^{2} â€“ 3abx + 2b^{2} and RHS = 0

â‰ RHS

And, for x = b/a

= RHS

Therefore, x = b/a is a solution of the given equation.