Question

find the equation whose roots are less than those of 2x³-7x²+3x-5=0 by 2​

Answer 1

Answer:

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