Question

12) The standard form of quadratic equation x - $ = 3x - 7 is
a)2x? - 8x + 7 = 0 b)2x2 – 7x+ 5 = 0
c)2x2 + 7x +5= 0 d)2x²–5x+ 7 = 0​

Answer 1

Step-by-step explanation:

Check whether the following are quadratic equations:

(i) (x + 1)2 = 2(x – 3)

(ii) x2 – 2x = (–2)(3 – x)

(iii) (x – 2)(x + 1) = (x – 1)(x + 3)

(iv) (x – 3)(2x + 1) = x(x + 5)

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

(vi) x2 + 3x +1 = (x – 2)2

(vii) (x + 2)3 = 2x(x2 – 1)

(viii) x3 – 4x2 – × + 1 = (x – 2)3

Sol. (i) (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 + 5 = 0

⇒ x2 – 11x + 8 = 0

Since x2 – 11x + 8 is a quadratic polynomial

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

(vi) x2 + 3x + 1 = (x – 2)2

We have:

x2 + 3x + 1 = (x – 2)2

⇒ x2 + 3x + 1 = x2 – 4x + 4

⇒ x2 + 3x + 1 – x2 + 4x – 4 =0

⇒ 7x – 3 = 0

Since 7x – 3 is a linear polynomial.

∴ x2 + 3x + 1 = (x – 2)2 is not a quadratic equation.

(vii) (x + 2)3 = 2x(x2 – 1)

We have:

(x + 2)3 = 2x(x2 – 1)

x3 + 3x2(2) + 3x(2)2 + (2)3 = 2x3 – 2x

⇒ x3 + 6x2 + 12x + 8 = 2x3 – 2x

⇒ x3 + 6x2 + 12x + 8 – 2x3 + 2x = 0

⇒ –x3 + 6x2 + 14x + 8 = 0

Since –x3 + 6x2 + 14x + 8 is a polynomial of degree 3

∴ (x + 2)3 = 2x(x2 – 1) is not a quadratic equation.

(viii) x3 – 4x2 – x + 1 = (x – 2)3

We have:

x3 – 4x2 – x + 1 = (x – 2)3

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

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

⇒ x3 – 4x2 – x – 1 – x3 + 6x2 – 12x + 8 = 0

2x2 – 13x + 9 = 0