Chapter 4 - Quadratic Equations, RD Sharma Solutions - (Part - 4) | RD Sharma Solutions for Class 10 Mathematics PDF Download

Page No 4.32

Q.3. Solve for x:



Ans.
(i) We have been given,

Now we solve the above equation as follows,

6x² - 30x + 30 = 10x² - 60x + 80
4x² - 30x + 50 = 0
2x² - 15x + 25 = 0
Now we also know that for an equation ax² + bx + c = 0, the discriminant is given by the following equation:
D = b² - 4ac
Now, according to the equation given to us, we have,a = 2, b = -15 and c = 25.
Therefore, the discriminant is given as,
D = (-15)² - 4(2)(25)
= 225 - 200
= 25
Now, the roots of an equation is given by the following equation,

Therefore, the roots of the equation are given as follows,

Now we solve both cases for the two values of x. So, we have,
x=(15+5)/4
= 5
Also,
x=(15-5)/4
=5/2
Therefore, the value of x = 5, 5/2.

(ii) We have been given,

Now we solve the above equation as follows,

-2 = 3x² - 6x
3x² - 6x + 2 = 0
Now we also know that for an equation ax² + bx + c = 0, the discriminant is given by the following equation:
D = b² - 4ac
Now, according to the equation given to us, we have, a = 3, b = -6 and c = 2.
Therefore, the discriminant is given as,

D = (-6)² - 4(3)(2)
= 36 - 24
= 12
Now, the roots of an equation is given by the following equation,

Therefore, the roots of the equation are given as follows,

Now we solve both cases for the two values of x. So, we have,

Also,

Therefore, the value of

(iii) We have been given,

Now, we solve the equation as follows:

x² + 1 = 3x
x² - 3x + 1 = 0
Now we also know that for an equation ax² + bx + c = 0, the discriminant is given by the following equation:
D = b² - 4ac
Now, according to the equation given to us, we have, a = 1, b = -3 and c = 1.
Therefore, the discriminant is given as,
D = (-3)² - 4(1)(1)
= 9 - 4
= 5
Now, the roots of an equation is given by the following equation,

Therefore, the roots of the equation are given as follows,

Now we solve both cases for the two values of x. So, we have,

Also,

Therefore, the value of

(iv) We have been given,

Now we solve the above equation as follows,

⇒ (16−x)(x+1)=15x
⇒ 16x+16−x²−x=15x
⇒ 15x+16−x²−15x=0
⇒ 16−x²=0
⇒ x²−16=0
Now we also know that for an equation ax² + bx + c = 0, the discriminant is given by the following equation:
D = b² - 4ac
Now, according to the equation given to us, we have, a = 1, b = -3 and c = 1.
Therefore, the discriminant is given as,
D=(0)²−4(1)(−16)
= 64
Now, the roots of an equation is given by the following equation,

Therefore, the roots of the equation are given as follows,

= ± 4
Therefore, the value of x = ± 4

(v) We have been given,


⇒ 48=x²+2x−15
⇒ x²+2x−15−48 = 0
⇒ x²+2x−63=0
⇒ x²+9x−7x−63 = 0
⇒ x(x+9)−7(x+9) = 0
⇒ (x−7)(x+9) = 0
⇒ x = 7, −9

Page No 4.4

Q.1. Which of the following are quadratic equations?
(i) x² + 6x − 4 = 0 
(ii) √3x² - 2x + 1/2 = 0
(iii) x² _{+ 1/x}² _{= 5}
(iv) x - 3/x = x²
(v) 2x² - √3x + 9 = 0
(vi) x² - 2x - √x - 5 = 0
(vii) 3x² - 5x + 9 = x² - 7x + 3
(viii) x + 1/x = 1
(ix) x² - 3x = 0
(x) (x + 1/x)²  = 3 (x + 1/x) + 4
(xi) (2x+1) (3x+2) = 6(x-1)(x-2)
(xii)  x + 1/x = x²_, x ≠ 0 
(xiii) 16x² − 3 = (2x + 5) (5x − 3)
(xiv) (x + 2)³ = x³ − 4
(xv) x(x + 1) + 8 = (x + 2) (x − 2)
Ans.

(i) Here it has been given that,
x² + 6x - 4 = 0
Now, the above equation clearly represents a quadratic equation of the form ax² + bx + c = 0, where a = 1, b = 6 and c = -4.
Hence, the above equation is a quadratic equation.

(ii) Here it has been given that,

Now, solving the above equation further we get,

Now, the above equation clearly represents a quadratic equation of the form ax² + bx + c = 0, where a = 2√3, b = -4 and c = 1.
Hence, the above equation is a quadratic equation.

(iii) Here it has been given that,

Now, solving the above equation further we get,

⇒ x⁴ + 1 = 5x²

⇒ x⁴ − 5x² + 1 = 0

Now, the above equation clearly does not represent a quadratic equation of the form ax² + bx + c = 0, because x⁴ − 5x² + 1 = 0 is a polynomial of degree 4.

Hence, the above equation is not a quadratic equation.

(iv) Here it has been given that,

Here it has been given that,

Now, solving the above equation further we get,

x² - 3 = x³
-x³ + x² - 3 = 0
Now, the above equation clearly does not represent a quadratic equation of the form ax² + bx + c = 0, because -x³ + x² - 3 = 0 is a polynomial of degree 3.
Hence, the above equation is not a quadratic equation.

(v) Here it has been given that,
_2x² _{- √3x + 9 = 0}
Now, the above equation clearly does not represent a quadratic equation of the form ax² + bx + c = 0, because 2x²-_√3x+9=0 contains a term x^1/2, where 1/2 is not an integer.
Hence, the above equation is not a quadratic equation.

(vi) Here it has been given that,
x² - 2x - √x - 5 = 0
Now, the above equation clearly does not represent a quadratic equation of the form ax² + bx + c = 0, because x² - 2x - √x - 5 = 0 contains an extra term x^1/2, where 1/2 is not an integer.
Hence, the above equation is not a quadratic equation.

(vii) Here it has been given that,
3x² - 5x + 9 = x² - 7x + 3
Now, after solving the above equation further we get,
2x²+2x+6 = 0
x² +x +3 = 0
Now, the above equation clearly does not represent a quadratic equation of the form ax² + bx + c = 0, where a = 1, b = 1 and c = 3. Hence, the above equation is a quadratic equation.

(viii) Here it has been given that,
x + 1/x = 1

Now, solving the above equation further we get,

x² + 1 = x
x² - x + 1 = 0
Now as we can see, the above equation clearly represents a quadratic equation of the form ax² + bx + c = 0, where a = 1, b = -1 and c = 1.
Hence, the above equation is a quadratic equation.

(ix) Here it has been given that,
 x² - 3x = 0
Now as we can see, the above equation clearly represents a quadratic equation of the form ax² + bx + c = 0, where a = 1, b = -3 and c = 0.
Hence, the above equation is a quadratic equation.

(x) Here it has been given that,

Now, solving the above equation further we get,

Now as we can see, the above equation clearly does not represent a quadratic equation of the form ax² + bx + c = 0, because x⁴ - 3x³ - 2x² - x + 1 is a polynomial having a degree of 4 which is never present in a quadratic polynomial.
Hence, the above equation is not a quadratic equation.

(xi) Here it has been given that,
(2x+1) (3x+2) = 6(x-1)(x-2)
Now, after solving the above equation further we get,
6x² +7x+ 2 =6x²-18x+12
25x-10=0
5x-2=0
Now as we can see, the above equation clearly does not represent a quadratic equation of the form ax² + bx + c = 0, because 5x - 2 = 0 is a linear equation.
Hence, the above equation is not a quadratic equation.
(xii) Here it has been given that,

Now, solving the above equation further we get,

x²+1 = 0
-x³+x²+1 = 0
Now as we can see, the above equation clearly does not represent a quadratic equation of the form ax² + bx + c = 0, because -x³+x²+1 = 0 is a polynomial having a degree of 3 which is never present in a quadratic polynomial.
Hence, the above equation is not a quadratic equation.

(xiii) Here it has been given that,
16x² − 3 = (2x + 5) (5x − 3)
Now, after solving the above equation further we get,
16x²-3  = 10x² + 19x - 15
4x² - 19x + 12 = 0
Now as we can see, the above equation clearly represents a quadratic equation of the form ax² + bx + c = 0, where a = 4, b = -19 and c = 12. Hence, the above equation is a quadratic equation.

(xiv) Here it has been given that,
(x + 2)³ = x³ − 4
Now, after solving the above equation further we get,
x³ +8 + 3(x)(2)(x+ 2) = x³ -4
12+6x² +12x = 0
x² + 2x + 2 = 0
Now as we can see, the above equation clearly represents a quadratic equation of the form ax² + bx + c = 0, where a= 1, b = 2 and c = 2.
Hence, the above equation is a quadratic equation.
(xv) Here it has been given that,
x(x + 1) + 8 = (x + 2) (x − 2)
Now, solving the above equation further we get,
x²+x+8 = x² - 4
x+12 = 0
Now as we can see, the above equation clearly does not represent a quadratic equation of the form ax² + bx + c = 0, because x+12 = 0 is a linear equation which does not have a x² term in it.

Q.2. In each of the following, determine whether the given values are solutions of the given equation or not:
(i) x²−3x+2=0, x=2, x=−1
(ii) x²+x+1=0, x=0, x=1
(iii) x²−3√3x+6=0, x=√3, x=−2√3
(iv) x+1/x=13/6, x=5/6, x=4/3
(v) 2x²−x+9=x²+4x+3, x=2, x=3
(vi) x²−√2x−4=0, x=−√2, x=−2√2
(vii) a²x²−3abx+2b²=0, x=a/b, x=b/a
Ans.
We are given the following quadratic equations and we are asked to find whether the given values are solutions or not
(i)  We have been given that,
x²−3x+2=0, x=2, x=−1
Now if x = 2 is a solution of the equation then it should satisfy the equation
So, substituting x = 2 in the equation we get
x² -3x + 2 = (2)² -3(2) + 2
= 4 - 6 + 2
= 0
Hence, x = 2 is a solution of the given quadratic equation.
Also, if x = -1 is a solution of the equation then it should satisfy the equation
So, substituting  x = -1 in the equation, we get
x² - 3x+3 = (-1)² - 3(-1) + 2

= 1 + 3+2
=6
Hence x = -1 is not a solution of the quadratic equation
Therefore, from the above results we find out that x = 2 is a solution and x = -1 is not a solution of the given quadratic equation.

(ii) We have been given that,
x²+x+1=0, x=0, x=1

Now if x = 0 is a solution of the equation then it should satisfy the equation.
So, substituting x = 0 in the equation, we get
x²+x+1 = (0)² + (1) + 1
= 1
Hence x = 0 is not a solution of the given quadratic equation.
Also, if x =1 is a solution of the equation then it should satisfy the equation.
So, substituting x = 0,  in the equation, we get
x²+x+1 = (1)² + (1) + 1
= 3
Hence x = 3 is not a solution of the quadratic equation.
Therefore, from the above results we find out that both x = 0 and x = -1 are not a solution of the given quadratic equation.