Question

(x+1) (3x+1) =(6x+2) (x+1) find the quadratic equation​

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

Step-by-step explanation:

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Answer 2

To solve this question,Let us first understand how to approach the solution by some examples :-

m( m² + n) = m³ + mn

x(x³ + x²) = x⁴ + x³

(a + b) (c + d) = a(c + d) + b( c + d) = ac + ad + bc + bd

Coming back to the questions :-

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

=> x( 3x + 1) + 1(3x + 1) = 6x(x + 1) + 2(x + 1)

=> 3x² + x + 3x + 1 = 6x² + 6x + 2x + 2

=> 6x² + 6x + 2x + 2 - 3x² - x - 3x - 1 = 0

=> 3x² + 4x + 1 = 0

Hence, 3x² + 4x + 1 = 0 is the required quadratic equation.

Finding roots of this quadratic equation by factorisation method :-

=> 3x² + 4x + 1 = 0

Process :- Split the middle term in such a way that the product of their coefficient equals the product of coefficient of the x² and the constant term.

=> 3x² + 3x + x + 1 = 0

=> 3x( x + 1) + 1(x + 1) = 0

=> (x + 1) (3x + 1) = 0

Hence, either x + 1 = 0 or 3x + 1 = 0

So, x = -1 or x = -1/3.