Q. 14.3( 204 Votes )

(i)

(ii)

(iii)

(iv)

(v)

Answer :

(i) X^{2} – 3x – 10

= x^{2} – 5x +2x – 10

= x(x – 5 ) + 2(x – 5 )

= (x – 5 )(x+2)

Roots of this equation are the values for which (x – 5 )(x+2) = 0

∴ x - 5 = 0 or x + 2 = 0

i.e.,

x = 5 or x = -2

(ii) 2 x^{2} + x – 6

= 2x^{2} + 4x – 3x – 6

= 2x (x+2) – 3(x+2)

= (x+2)(2x – 3 )

Roots of this equation are the values for which (x+2)(2x – 3) = 0

∴ x+ 2= 0 or 2x – 3 = 0

*x* = −2 or *x* =

(iii)

Roots of this equation are the values for which = 0

∴ = 0 or = 0

(iv)

Roots of this equation are the values for which (4x – 1 )^{2} = 0

Therefore,

(4x – 1 ) = 0 or (4x – 1) = 0

i.e.,

(v) 100x^{2} – 20x +1

= 100x^{2} – 10x – 10x +1

=10x(10x – 1) – 1 (10x – 1)

=(10x – 1)^{2}

Roots of this equation are the values for which (10x – 1)^{2} = 0

Therefore,

(10x – 1) = 0 or (10x – 1) = 0

i.e.,

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