Question

Q:-factorise and hence find roots
${x}^{2} + {(x + 1) }^{2} = 365$
​

Answer 1

Answer:

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QUESTION:−

Q:-factorise and hence find roots

{x}^{2} + {(x + 1) }^{2} = 365x

2

+(x+1)

2

=365

\huge\tt\underline\blue{Answer }

Answer

_ _ _ _ _ _ _ _ _ _ _ _ _ _ _✍️

⟹ {x}^{2} + {(x + 1)}^{2} = 365⟹x

2

+(x+1)

2

=365

⟹ {x}^{2} + {x}^{2} + 2x + 1 = 365⟹x

2

+x

2

+2x+1=365

⟹2 {x}^{2} + 2x - 364 = 0⟹2x

2

+2x−364=0

⟹ {x}^{2} + x - 182 = 0⟹x

2

+x−182=0

⟹ {x}^{2} + 14x - 13x - 182 = 0⟹x

2

+14x−13x−182=0

⟹x(x + 14) - 13(x + 14) = 0⟹x(x+14)−13(x+14)=0

⟹(x + 14)(x - 13) = 0⟹(x+14)(x−13)=0

⟹x + 14 = 0⟹x+14=0

⟹x - 13 = 0⟹x−13=0

⟹x = - 14⟹x=−14 ✓

⟹x = 13⟹x=13 ✓

HOPE IT HELPS YOU..

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Answer 2

Answer:

x + 1)2 = 2(x – 3)

We have:

(x + 1)2 = 2 (x – 3) x2 + 2x + 1 = 2x – 6

⇒ x2 + 2x + 1 – 2x + 6 = 0

⇒ x2 + 70

Since x2 + 7 is a quadratic polynomial

∴ (x + 1)2 = 2(x – 3) is a quadratic equation.

(ii) x2– 2x = (–2) (3 – x)

We have:

x2 – 2x = (– 2) (3 – x)

⇒ x2 – 2x = –6 + 2x

⇒ x2 – 2x – 2x + 6 = 0

⇒ x2 – 4x + 6 = 0

Since x2 – 4x + 6 is a quadratic polynomial

∴ x2 – 2x = (–2) (3 – x) is a quadratic equation.

(iii) (x – 2) (x + 1) = (x – 1) (x + 3)

We have:

(x – 2) (x + 1) = (x – 1) (x + 3)

⇒ x2 – x – 2 = x2 + 2x – 3

⇒ x2 – x – 2 – x2 – 2x + 3 = 0

⇒ –3x + 1 = 0

Since –3x + 1 is a linear polynomial

∴ (x – 2) (x + 1) = (x – 1) (x + 3) is not quadratic equation.

(iv) (x – 3) (2x + 1) = x(x + 5)

We have:

(x – 3) (2x + 1) = x(x + 5)

⇒ 2x2 + x – 6x – 3 = x2 + 5x

⇒ 2x2 – 5x – 3 – x2 – 5x – 0

⇒ x2 + 10x – 3 = 0

Since x2 + 10x – 3 is a quadratic polynomial

∴ (x – 3) (2x + 1) = x(x + 5) is a quadratic equation.

(v) (2x – 1) (x – 3) = (x + 5) (x – 1)

We have:

(2x – 1) (x – 3) = (x + 5) (x – 1)

⇒ 2x2 – 6x – x + 3 = x2 – x + 5x – 5

⇒ 2x2 – x2 – 6x – x + x – 5x + 3 +