Ex-8.4 Quadratic Equations, Class 10, Maths RD Sharma Solutions | Extra Documents, Videos & Tests for Class 10 PDF Download

By using the method of completing the square, find the roots of the following quadratic equations: (1)  x² –  + 6 = 0

Sol:

x² –  + 6 = 0

i.e. x² – 2×x×2√2 + (2√2)² – (2√2)² + 6 = 0

(x – 2√2)² = (2√2)² – 6

= (x – 2√2)² = (4×2) – 6

=  (x – 2√2)² = 8 – 6

=  (x – 2√2)² = 2

=  (x – 2√2)  =  ±√2

=  (x – 2√2)  =  √2  or  (x – 2√2)  =   –  √2

x = √2 + 2√2  or    x = – √2 + 2√2

=  x = √3/2  or  x = √2

So, the roots for the given equation are :

x = √3/2  or  x = √2

(2)  2x² – 7x + 3 = 0

Sol:

2x² – 7x + 3 = 0

2(x² – 7x/2 + 3/2) = 0

x² – 2×7/2×1/2×x + 3/2 = 0

x² – 2×74×x + (7/4)² – (7/4)² + 3/2 = 0

x² – 2×7/4×x + (7/4)² – (49/16) + 3/2 = 0

(x – 7/4)² – 49/16 + 3/2 = 0

(x – 7/4)² = 49/16 – 3/2

(x – 7/4)² = (49 – 26)/16

(x – 7/4)² = 25/16

(x – 7/4)² = (5/4)²

x – 7/4 = ±5/4

x – 7/4 = 5/4  or  x – 7/4 = – 5/4

x = 7/4 + 5/4  or  x = 7/4 – 5/4

x = 12/4  or  x = 2/4

x = 3  or  x = 1/2

(3)  3x² + 11x + 10 = 0

Sol:  3x² + 11x + 10 = 0

x² + 11x/3 + 10/3 = 0

x² + 2 × 1/2 × 11x/3 + 10/3 = 0

x² + 2 × 11x/6 + (11/6)² – (11/6)² + 10/3 = 0

(x + 11/6)² = (11/6)² – 10/3

(x + 11/6)² = 121/36 – 10/3

(x + 11/6)² = (121 – 120)/36

(x + 11/6)² = 136

(x + 11/6)² = (16)²

x + 11/6 = ±1/6

x + 11/6 = 1/6  or  x + 11/6 = – 1/6

x = 1/6 – 11/6  or  x = – 1/6 – 11/6

x = – 10/6  or  x = – 12/6  = – 2

x = – 5/3   or  x = – 2

(4)  2x² + x – 4 = 0

Sol:  2x² + x – 4 = 0

2(x² + x/2 – 4/2) = 0

x² + 2×1/2×1/2×x – 2 = 0

x² + 2×1/4×x + (1/4)² – (1/4)² – 2 = 0

(x + 1/4)² = (1/4)² + 2

(x + 1/4)²  =  (1/4)² + 2

(x + 1/4)²  =  11/6 + 2

(x + 1/4)²  =  1 + 2×16/16

(x + 1/4)²  =  1 + 32/16

(x + 1/4)²  =  33/16

(x + 14)  =  ±

So, 

Are the two roots of the given equation.

(5)  2x² + x + 4 = 0

Sol:  2x² + x + 4 = 0

x² + x/2 + 2 = 0

x² + 2×1/2×1/2×x + 2 = 0

x² + 2×1/4×x + (1/4)² – (1/4)² + 2 = 0

x² + 2×14×x + (1/4)² = (1/4)² – 2

(x + 1/4)² = 11/6 – 2

Since,    is not a real number,

Therefore, the equation doesn’t have real roots.

(6)  4x² + 4√3x + 3 = 0

Sol:  4x² + 4√3x + 3 = 0

x² +  + 3/4 = 0

x² + 2 × 1/2 × √3 × x + 34 = 0

x² + 2×√3/2×x + (√3/2)² – (√3/2)² + 34 = 0

(x + √3/2)² – 3/4 + 3/4 = 0

(x + √3/2)²   = 0

(x + √3/2)  = 0  and  (x + √3/2)  = 0

x = – √3/2  and x = – √3/2

Therefore, x = – √3/2  and x = – √3/2

Are the real roots of the given equation.

(7)  √2x² – 3x – 2√2 = 0

Sol:   √2x² – 3x – 2√2 = 0

(8)  √3x² + 10x + 7√3 = 0

Sol:  √3x² + 10x + 7√3 = 0

x² + 2×1/2×10x/√3 + 7 = 0

(x + 5/√3)² = 25/3 – 7

(x + 5/√3)² = (25 – 21)/3

(x + 5/√3)² = 4/3

x + 5/√3 = + 2/√3 or x + 5/√3 = – 2√3

x = – 3/√3 or x = – 7√3

x = – √3 and x = – 7/√3

(9)  x² – (√2 + 1)x + √2 = 0

Sol: x² – (√2 + 1)x + √2 = 0

x² – 2×1/2(√2 + 1)x + √2 = 0

(10)  x²  – 4ax + 4a²  – b²  = 0

Sol:   x²  – 4ax + 4a²  – b²  = 0

x²  – 2(2a).x + (2a)²  – b²  = 0

(x – 2a)²  = b²

x – 2a =  ± b

x – 2a = b or x – 2a = – b

x = 2a + b or  x = 2a – b

Therefore, x = 2a + b or  x = 2a – b are the two roots of the given equation.