RS Aggarwal Solutions Class 10 Chapter 10 Quadratic Equations 10A

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)   x²-x+3=0 is a quadratic polynomial.
∴  x²-x+3=0 is a quadratic equation.

(ii) 2x²+ \(\frac { 5 }{ 2 }   \)x-√3=0
⇒ 4x²+5x-2√3=0
Clearly is 4x²+5x-2√3=0 a quadratic polynomial.
∴ 2x²+ \(\frac { 5 }{ 2 }   \)x-√3=0 is a quadratic equation.

(iii) √2x²+7x+5√2=0 is a quadratic polynomial.
∴ √2x²+7x+5√2=0 is a quadratic equation.

(iv)\(\frac { 1 }{ 3 }   \)x²+\(\frac { 1 }{ 5 }   \)x-2=0
⇒ 5x²+3x-2=0
Clearly, 5x²+3x-2=0 is a quadratic equation.
\(\frac { 1 }{ 3 }   \)x²+\(\frac { 1 }{ 5 }   \) is a quadratic equation.

(v) x²-3x-√x+4=0 is not a quadratic polynomial since it contains √x, in which power 1/2 of x is not an integer.
∴ x²-3x-√x+4=0 is not a quadratic equation.

(vi) x-\(\frac { 6}{ x }   \)=3
⇒ x²-3x-6 =0
And (x²-3x-6)Being a polynomial of degree 2, it is a quadratic polynomial.
Hence, x-\(\frac { 6}{ x }   \)=3 is a quadratic equation.

(vii) x+\(\frac { 2}{ x }   \)= x²
⇒ x³-x²-2 =0
And (x³-x²-2 =0) being a polynomial of degree 3, it is not a quadratic polynomial.
Hence, x+\(\frac { 2}{ x }   \)= x² is not a quadratic equation.

(viii) \({ x }^{ 2 }-\frac { 1 }{ { x }^{ 2 } } =5  \) ⇒ x⁴ -1=5x²
⇒x⁴-5x²-1 =0
And (x⁴-5x²-1 =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 3x²+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

• RS Aggarwal Solutions Class 10
• RD Sharma Class 10 Solutions
• NCERT Solutions

Question 3:
Since x = 1 is a solution of x²+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  ax²+bx-6=0, we have


Again x = -2 being a root of ax²+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 3x²-243=0.
Question 8:

Hence, -5 and -7 are the roots of x²+12x+35=0.
Question 9:

Hence, 11 and 7 are the roots of equation x²=18x-77
Question 10:

Hence, \(x=-\frac { 1 }{ 3 }    \) is the repeated root of the equation 9x²+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 6x²+11x+3=0
Question 13:

Hence, \(x=\frac { 4 }{ 3 }   \) and \(x=-\frac { 3 }{ 2 }   \) are the roots of equation 6x²+x-12=0

Question 14:

Hence, \(x=-\frac { 1 }{ 3 }   \) and 1 are the roots of the equation 3x²-2x-1=0.
Question 15:

Hence, \(x=\frac { 2 }{ 3 }   \) and \(x=-\frac { 1 }{ 2 }   \)are the roots of equation 6x²-x-2=0.
Question 16:

Hence, \(x=-\frac { 1 }{ 16 }   \) and \(x=\frac { 2 }{ 3 }   \) are the roots of 48x²-13x-1=0.
Question 17:

Hence, \(x=-\frac { 5 }{ 3 }   \) and x=-2 are the roots of the equation 3x²+11x+10=0
Question 18:

Hence,\(x=\frac { 25 }{ 4 }   \) and x=-4 are the roots of the equation 4x²-9x=100.
Question 19:

Hence,  \(x=\frac { 4 }{ 9 }   \) and 2 are the roots of the equation 9x²-22+8=0
Question 20:

Hence,  \(x=\frac { 7 }{ 5 }   \) and \(x=-\frac { 4 }{ 3 }   \) are the roots of the given equation 15x²-28=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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