# 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

### 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

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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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