## RS Aggarwal Solutions Class 10 Chapter 10 Quadratic Equations 10A

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 10 Quadratic Equations 10A

**Other Exercises**

- RS Aggarwal Solutions Class 10 Chapter 10 Quadratic Equations 10A
- RS Aggarwal Solutions Class 10 Chapter 10 Quadratic Equations 10B
- RS Aggarwal Solutions Class 10 Chapter 10 Quadratic Equations 10C
- RS Aggarwal Solutions Class 10 Chapter 10 Quadratic Equations 10D

**Exercise 10A**

**Question 1:**

(i) x^{2}-x+3=0 is a quadratic polynomial.

∴ x^{2}-x+3=0 is a quadratic equation.

(ii) 2x^{2}+ \(\frac { 5 }{ 2 } \)x-√3=0

⇒ 4x^{2}+5x-2√3=0

Clearly is 4x^{2}+5x-2√3=0 a quadratic polynomial.

∴ 2x^{2}+ \(\frac { 5 }{ 2 } \)x-√3=0 is a quadratic equation.

(iii) √2x^{2}+7x+5√2=0 is a quadratic polynomial.

∴ √2x^{2}+7x+5√2=0 is a quadratic equation.

(iv)\(\frac { 1 }{ 3 } \)x^{2}+\(\frac { 1 }{ 5 } \)x-2=0

⇒ 5x^{2}+3x-2=0

Clearly, 5x^{2}+3x-2=0 is a quadratic equation.

\(\frac { 1 }{ 3 } \)x^{2}+\(\frac { 1 }{ 5 } \) is a quadratic equation.

(v) x^{2}-3x-√x+4=0 is not a quadratic polynomial since it contains √x, in which power 1/2 of x is not an integer.

∴ x^{2}-3x-√x+4=0 is not a quadratic equation.

(vi) x-\(\frac { 6}{ x } \)=3

⇒ x^{2}-3x-6 =0

And (x^{2}-3x-6)Being a polynomial of degree 2, it is a quadratic polynomial.

Hence, x-\(\frac { 6}{ x } \)=3 is a quadratic equation.

(vii) x+\(\frac { 2}{ x } \)= x^{2}

⇒ x^{3}-x^{2}-2 =0

And (x^{3}-x^{2}-2 =0) being a polynomial of degree 3, it is not a quadratic polynomial.

Hence, x+\(\frac { 2}{ x } \)= x^{2} is not a quadratic equation.

(viii) \({ x }^{ 2 }-\frac { 1 }{ { x }^{ 2 } } =5 \) ⇒ x^{4} -1=5x^{2}

⇒x^{4}-5x^{2}-1 =0

And (x^{4}-5x^{2}-1 =0) being a polynomial of degree 4.

Hence \({ x }^{ 2 }-\frac { 1 }{ { x }^{ 2 } } =5 \) is not a quadratic equation.

Solving A Quadratic Equation By Completing The Square

**Question 2:**

The given equation is 3x^{2}+2x-1=0

(i) On substituting x = -1 in the equation, we get

(ii) On substituting \(x=\frac { 1 }{ 3 } \) in the equation, we get

(iii) On substituting \(x=-\frac { 1 }{ 2 } \) in the equation , we get

**More Resources**

**Question 3:**

Since x = 1 is a solution of x^{2}+kx+3=0 it must satisfy the equation.

Hence the required value of k = -4

**Question 4:**

Since \(x=\frac { 3 }{ 4 } \) is a root of ax^{2}+bx-6=0, we have

Again x = -2 being a root of ax^{2}+bx-6=0, we have

Multiplying (2) by 4 adding the result from (1), we get

11a = 44 ⇒ a = 4

Putting a = 4 in (1), we get

**Question 5:**

**Question 6:**

**Question 7:**

Hence, 9 and -9 are the roots of the equation 3x^{2}-243=0.

**Question 8:**

Hence, -5 and -7 are the roots of x^{2}+12x+35=0.

**Question 9:**

Hence, 11 and 7 are the roots of equation x^{2}=18x-77

**Question 10:**

Hence, \(x=-\frac { 1 }{ 3 } \) is the repeated root of the equation 9x^{2}+6x+1=0

**Question 11:**

Hence, is the repeated root of the equation

**Question 12:**

Hence, \(x=-\frac { 3 }{ 2 } \), \(x=-\frac { 1 }{ 2 } \)are the roots of 6x^{2}+11x+3=0

**Question 13:**

Hence, \(x=\frac { 4 }{ 3 } \) and \(x=-\frac { 3 }{ 2 } \) are the roots of equation 6x^{2}+x-12=0

**Question 14:**

Hence, \(x=-\frac { 1 }{ 3 } \) and 1 are the roots of the equation 3x^{2}-2x-1=0.

**Question 15:**

Hence, \(x=\frac { 2 }{ 3 } \) and \(x=-\frac { 1 }{ 2 } \)are the roots of equation 6x^{2}-x-2=0.

**Question 16:**

Hence, \(x=-\frac { 1 }{ 16 } \) and \(x=\frac { 2 }{ 3 } \) are the roots of 48x^{2}-13x-1=0.

**Question 17:**

Hence, \(x=-\frac { 5 }{ 3 } \) and x=-2 are the roots of the equation 3x^{2}+11x+10=0

**Question 18:**

Hence,\(x=\frac { 25 }{ 4 } \) and x=-4 are the roots of the equation 4x^{2}-9x=100.

**Question 19:**

Hence, \(x=\frac { 4 }{ 9 } \) and 2 are the roots of the equation 9x^{2}-22+8=0

**Question 20:**

Hence, \(x=\frac { 7 }{ 5 } \) and \(x=-\frac { 4 }{ 3 } \) are the roots of the given equation 15x^{2}-28=x.

**Question 21:**

Hence, \(x=\frac { 1 }{ 3 } \) and -4 are the roots of given equation .

**Question 22:**

Hence, 1 and √2 are the roots of the given equation

**Question 23:**

**Question 24:**

**Question 25:**

Hence, \(\frac { -\sqrt { 7 } }{ 3 } \) and \(\frac { \sqrt { 7 } }{ 7 } \) are the roots of given equation.

**Question 26:**

Hence, -√7 and \(\frac { 13\sqrt { 7 } }{ 7 } \) are the roots of given equation.

**Question 27:**

Hence, \(\frac { 2\sqrt { 6 } }{ 3 } \) and \(\frac { -\sqrt { 6 } }{ 8 } \)are the roots of given equation.

**Question 28:**

Hence, 5 and \(-\frac { 7 }{ 5 } \)are the roots of given equation

**Question 29:**

Hence, \(-\frac { 1 }{ 5 } \) and \(\frac { 1 }{ 2 } \)are the roots of given equation.

**Question 30:**

Hence, 2 and \(\frac { 1 }{ 2 } \) are the roots of given equation.

**Question 31:**

Hence, \(-\frac { b }{ a } \) and \(\frac { c }{ b } \) are the roots of given equation.

**Question 32:**

Hence, \(\frac { -1 }{ { a }^{ 2 } } \) and \(\frac { 1 }{ { b }^{ 2 } } \)are the roots of given equation.

**Question 33:**

Hence, \(\frac { 3a }{ 4b } \) and \(\frac { -2b }{ 3a } \) are the roots of given equation.

**Question 34:**

Hence, \(\frac { { a }^{ 2 } }{ 2 } \) and \(\frac { { b }^{ 2 } }{ 2 } \)are the roots of given equation.

**Question 35:**

Hence, 2 and 1 are the roots of the given equation

**Question 36:**

Hence, -9 and 7 are the roots of the given equation

**Question 37:**

Hence, -4 and \(\frac { 9 }{ 4 } \) are the roots of the given equation

**Question 38:**

Hence \(\frac { 40 }{ 13 } \) and 6 are the roots of the given equation

**Question 39:**

Hence, 4 and \(-\frac { 2 }{ 9 } \) are the roots of the given equation

**Question 40:**

Hence, 3 and \(\frac { 4 }{ 3 } \) are the roots of the given equation.

**Question 41:**

Hence, 5 and \(\frac { 1 }{ 2 } \) are the roots of the given equation.

**Question 42:**

Putting the given equation become

Case I:

Case II:

Hence, \(-\frac { 3 }{ 2 } \) and -2 are the roots of the given equation

**Question 43:**

Putting the given equation become

Case I:

Case II:

Hence, -1 and \(-\frac { 23 }{ 5 } \) are the roots of the given equation

**Question 44:**

On putting the given equation become

Case I:

Case II:

Hence, -10 and \(-\frac { 1 }{ 5 } \) are the roots of the given equation.

**Question 45:**

Putting the given equation become

Case I:

Case II:

Hence, -1 and \(\frac { 1 }{ 8 } \) are the roots of the given equation

**Question 46:**

The given equation

Hence, (a+b) and \(\frac { (a+b) }{ 2 } \) is the roots of the given equation

**Question 47:**

Hence, \(\frac { a+b }{ ab } \) and \(\frac { 2 }{ a+b } \) are the roots of the given equation

**Question 48:**

Hence, -2,0 are the roots of the given equation

**Question 49:**

Hence, \(\frac { 1 }{ 2 } \) and \(\frac { 1 }{ 2 } \) are the roots of the given equation

**Question 50:**

Hence, 3 and 2 are the roots of the given equation.

Hope given RS Aggarwal Solutions Class 10 Chapter 10 Quadratic Equations 10A are helpful to complete your math homework.

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