Math, asked by riyanadcunha15, 7 months ago

Class 10 CBSE Subject- Math Ch 4 Quadratic equation. Ex 4.1 solve Qn 1 all sub questions.Pls explain in detail.
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Answered by Aniketaslaliya
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Answered by Anonymous
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 \huge \underline \bold \orange{solution \:  - }

(i) (x+ 1)²=2(x-3)

⇒ x² + 2x +1 = 2x – 6

⇒ x² + 2x – 2x+1 + 6 = 0

⇒ x² + 7 = 0

⇒ x² + 0x + 1 = 0

Which is of the form

ax2 + bx + c = 0

Hence, the given equation is a quadratic equation.

(ii) x² – 2x = (- 2) (3 -x)

⇒ x² -2x = -6 + 2x

⇒ x² -4x + 6 = 0

Which is of the form

ax2 + bx + c = 0

Hence, the given equation is a quadratic equation.

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

⇒ x² + x-2x-2 = x² + 3x-x -3

⇒ x² + x – 2x – 2 = x² – 3x + x + 3 = 0

⇒ – 3x + 1 = 0 ⇒ 3x – 1 = 0

Since degree of equation is 1, hence, given equation is not a quadratic equation.

(iv) (x-3) (2x+ 1) = x (x + 5)

⇒ 2x² + x – 6x – 3 = x² + 5x

2x² + x – 6x – 3-x² – 5x = 0

⇒ x² – 10x -3 = 0

Which is of the form

ax² + bx + c = 0

Hence, the given equation is a quadratic equation.

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

⇒ 2x² – 6x-x + 3 = x² -x + 5x – 5

2x² – 6x-x + 3 = x² + x – 5x + 5 = 0

⇒ x² – 11x + 8 = 0

Which is of the form

ax² + bx + c = 0

Hence, the given equation is a quadratic equation.

(vi) x² + 3x + 1 = (x-2)²

⇒ x² + 3x + 1 = x² + 4 – 4x

⇒ x² + 3x + 1 = x²– 4 + 4c = 0

⇒ 7x – 3 = 0

Since degree of equation is 1, hence, the given equation is not a quadratic equation,

(vii) (x + 2)³ = 2x (x² – 1)

⇒ x³ + 8 + 3.x.2 (x + 2) = 2x³ – 2x

⇒ x³ + 8 + 6x² + 12x = 2x³ – 2x

⇒ x³ – 6x² – 14x – 8 = 0

Which is not of the form

ax2 + bx + c = 0

Hence, the given equation is not a quadratic equation.

(viii) x³ – 4x² – x+1 = (x-2)³

⇒ x³ – 4x² – x + 1 = x3-8 + 3x(-2)(x – 2)

⇒ x³ – 4x² -x + 1 = x³ – 6x² + 12x – 8

⇒ 2x² – 13x + 9 = 0

Which is of the form

ax2 + bx + c = 0

Hence, the given equation is a quadratic equation.

Thankyou....


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