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

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

**(i) x**^{2} + 6x – 4 = 0

^{2}+ 6x – 4 = 0

**Solution:**

As we know that, quadratic equation is in form of: ax

^{2}+ bx + c = 0Here, given equation is x

^{2}+ 6x – 4 =0 in quadratic equation.So, It is a quadratic equations.

**(ii) âˆš3x**^{2} âˆ’ 2x + Â½= 0

^{2}âˆ’ 2x + Â½= 0

**Solution:**

As we know that, quadratic equation is in form of: ax

^{2}+ bx + c = 0Here, given equation is âˆš3x

^{2}âˆ’ 2x + Â½= 0 in quadratic equation.So, It is a quadratic equation.

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

^{2}+ 1/x

^{2 }= 5

**Solution:**

This equation can be written as: x

^{4}– 5x^{2}+ 1 =0As we know that, quadratic equation is in form of: ax

^{2}+ bx + c = 0Here, given equation is x

^{4}– 5x^{2}+ 1 =0 is 4 degree polynomial.So, It is not quadratic equations.

**(iv) x – 3/x = x**^{2}

^{2}

**Solution:**

This equation can be written as: x

^{3}– x^{2}– 3 = 0As we know that, quadratic equation is in form of: ax

^{2}+ bx + c = 0Here, given equation is x

^{3}– x^{2}– 3 = 0 is 3 degree polynomial.So, It is not quadratic equations.

**(v) 2x**^{2} – âˆš3x + 9 = 0

^{2}– âˆš3x + 9 = 0

**Solution:**

As we know that, quadratic equation is in form of: ax

^{2}+ bx + c = 0Here, given equation is 2x

^{2 }– âˆš3x + 9 = 0 in quadratic equation.So, It is a quadratic equation.

**(vi) x**^{2} – 2x -âˆšx – 5 = 0

^{2}– 2x -âˆšx – 5 = 0

**Solution:**

As we know that, quadratic equation is in form of: ax

^{2}+ bx + c = 0Here, given equation is x

^{2}– 2x -âˆšx -5 = 0.So, It is not quadratic equation.

**(vii) 3x**^{2} âˆ’ 5x + 9 = x^{2} âˆ’ 7x + 3

^{2}âˆ’ 5x + 9 = x

^{2}âˆ’ 7x + 3

**Solution:**

This equation can be written as: 2x

^{2}+ 2x + 6 = 0As we know that, quadratic equation is in form of: ax

^{2}+ bx + c = 0Here, given equation is 2x

^{2}+ 2x + 6 = 0 in quadratic equation.So, It is a quadratic equation.

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

**Solution:**

This equation can be written as: x

^{2}– x + 1 = 0As we know that, quadratic equation is in form of: ax

^{2}+ bx + c = 0Here, given equation is x

^{2}– x + 1 = 0 in quadratic equation.So, It is a quadratic equation.

**(ix) x**^{2} – 3x = 0

^{2}– 3x = 0

**Solution:**

As we know that, quadratic equation is in form of: ax

^{2}+ bx + c = 0Here, given equation is x

^{2}– 3x =0 in quadratic equation.So, It is a quadratic equation.

**(x) (x + 1/x)**^{2} = 3(x +1/x)

^{2}= 3(x +1/x)

**Solution:**

((x

^{2}+ 1)/x)^{2 }= 3((x^{2 }+ 1)/2)This equation can be written as: x

^{5}– 3x^{4}+ 2x^{3 }– 3x^{2 }+ x =0As we know that, quadratic equation is in form of: ax

^{2}+ bx + c = 0Here, given equation is x

^{5}– 3x^{4}+ 2x^{3}– 3x^{2}+ x = 0 is 5 degree polynomial.So, It is not quadratic equations.

**(xi) (2x**^{2} + 1)(3x^{2} + 2) = 6(x^{2}âˆ’ 1)(x^{2}âˆ’ 2)

^{2}+ 1)(3x

^{2}+ 2) = 6(x

^{2}âˆ’ 1)(x

^{2}âˆ’ 2)

**Solution:**

This equation can be written as:

6x

^{2}+ 4x + 3x + 2 = 6x^{2}-12x – 6x + 127x + 2 = -18x + 12

This equation can be written as: 25x â€“ 10 = 0

As we know that , quadratic equation is in form of: ax

^{2}+ bx + c = 0Here, given equation is 25x â€“ 10 = 0 is 1 degree polynomial.

So, It is not quadratic equations.

**(xii) x + 1/x = x**^{2}, x â‰ ** 0**

^{2}, x

**Solution:**

x + 1/x = x

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

^{2}+ 1 = x^{3}This equation can be written as: x

^{3}â€“ x^{2}â€“ 1 = 0As we know that, quadratic equation is in form of: ax

^{2}+ bx + c = 0Here, given equation x

^{3}â€“ x^{2}â€“ 1 = 0 is = 0 is 3 degree polynomial.So, It is not quadratic equations.

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

^{2}â€“ 3 = (2x + 5)(5x – 3)

**Solution:**

16x

^{2}â€“ 3 = (2x + 5)(5x – 3)16x

^{2}â€“ 3 = 10x^{2}â€“ 6x + 25x â€“ 15This equation can be written as: 6x

^{2}â€“ 19x + 12 = 0As we know that , quadratic equation is in form of: ax

^{2}+ bx + c = 0Here, given equation is 6x

^{2}â€“ 19x + 12 = 0 is 2 degree polynomial.So, It is quadratic equations.

**(xiv) (x + 2)**^{3} = x^{3} â€“ 4

^{3}= x

^{3}â€“ 4

**Solution:**

(x + 2)

^{3}= x^{3}â€“ 4On expanding, we get x

^{3}+ 6x^{2}+ 8x + 8 = x^{3}â€“ 46x

^{2}+ 8x + 12 = 0This equation can be written as: 6x

^{2}+ 8x + 12 = 0As we know that, quadratic equation is in form of: ax

^{2}+ bx + c = 0Here, given equation is 6x

^{2}â€“ 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)

x

^{2}+ x + 8 = x^{2}â€“ 4x + 12 = 0

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

As we know that, quadratic equation is in form of: ax

^{2}+ bx + c = 0Here, given equation is x + 12 = 0 is 0 degree polynomial.

So, It is not a quadratic equation.

**Question 2. In each of the following, determine whether the given values are solutions of the given equation or not:**

**(i) x**^{2} – 3x + 2 = 0, x = 2, x = – 1

^{2}– 3x + 2 = 0, x = 2, x = – 1

**Solution:**

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

LHS = x

^{2}– 3x + 2Now we will put x = 2 in LHS, we get

(2)

^{2}â€“ 3(2) + 2â‡’ 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) + 21 + 3 + 2 = 6 â‰ RHS

LHS â‰ RHS

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

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

^{2}+ x + 1 = 0, x = 0, x = 1

**Solution:**

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

LHS = x

^{2}+ x + 1Now we will put x = 0 in LHS, we get

(0)

^{2}+ 0 + 1â‡’ 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 + 1â‡’ 3 â‰ RHS

â‡’ LHS â‰ RHS

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

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

^{2}âˆ’ 3âˆš3x + 6 = 0, x = âˆš3 and x = âˆ’2âˆš3

**Solution:**

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

LHS = x

^{2}âˆ’ 3âˆš3x + 6Now, we will put x = âˆš3 in LHS, we get

(âˆš3)

^{2}âˆ’ 3âˆš3(âˆš3) + 6â‡’ 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) + 6â‡’ 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) 2x**^{2} – x + 9 = x^{2}** + 4x + 3, x = 2, x = 3**

^{2}– x + 9 = x

**Solution:**

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

2x

^{2}– x + 9 = x^{2}+ 4x + 3â‡’ x

^{2}â€“ 5x + 6 = 0LHS = x

^{2}– 5x + 6Now, we will put x = 2 in LHS, we get

(2)

^{2}â€“ 5(2) + 64 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) + 6â‡’ 9 â€“ 15 + 6 = 0 = RHS

â‡’ LHS = RHS

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

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

^{2}â€“ âˆš2x â€“ 4 = 0, x = -âˆš2, x = -2âˆš2

**Solution:**

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

LHS = x

^{2}â€“ âˆš2x â€“ 4Now, we will put x = -âˆš2 in LHS, we get

(-âˆš2)

^{2}âˆ’ âˆš2(-âˆš2) â€“ 4â‡’ 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) – 4â‡’ 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) a**^{2}x^{2} â€“ 3abx + 2b^{2} = 0, x = a/b, x = b/a

^{2}x

^{2}â€“ 3abx + 2b

^{2}= 0, x = a/b, x = b/a

**Solution:**

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

LHS = a

^{2}x^{2}â€“ 3abx + 2b^{2}Now, we will put x = a/b in LHS, we get

a

^{2}(a/b)^{2}-3ab(a/b) + 2b^{2}â‡’ a

^{4}/b^{2}– 3a^{2}+ 2b^{2}â‰ RHSâ‡’ 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

a

^{2}(b/a)^{2}– 3ab(b/a) + 2b^{2}â‡’ b

^{2}– 3b^{2}+ 2b^{2}= RHSâ‡’ LHS = RHS

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

**Question 3: In each of the following, find the value of k for which the given value is a solution of the given equation.**

**(i) 7x**^{2} + kx – 3 = 0, x = 2/3

^{2}+ 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 =028/9 + 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) x**^{2} – x(a + b) + k = 0, x = a

^{2}– 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

a

^{2}– a^{2}– ab + k = 0k = ab

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

**(iii) kx**^{2} + âˆš2x – 4 = 0, x = âˆš2

^{2}+ 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 = 02k + 2 â€“ 4 = 0

2k â€“ 2 = 0

k = 1

On putting value of x = âˆš2 , we get k = 1.

**(iv) x**^{2} + 3ax + k = 0, x = -a

^{2}+ 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 = 0a

^{2}– 3a^{2}+ k = 0k = 2a

^{2}On putting value of x = – a, we get k = 2a

^{2}.

**Question 4. Given to check whether 3 is a root of the equation**

**âˆš(x**^{2 }– 4x +3) + âˆš(x^{2} -9) = âˆš(4x^{2} – 14x + 16)

^{2 }– 4x +3) + âˆš(x

^{2}-9) = âˆš(4x

^{2}– 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 ax**^{2} + 7x + b = 0, find the values of a and b.

^{2}+ 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 =04a/9 + 14/3 + b =0

4a + 42 + 3b = 0 — (i)

Now, x = -3.

a(-3)

^{2}+ 7(-3) + b = 09a – 21 + b = 0 –(ii)

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

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