Question

find the root of the following quadratic equations by using the method of factorisation x+1 whole square- 16=0​

Answer 1

Answer:

(x+1)^2-16=0

or,x^2+2x-15=0

or,(x+5)(x-3)=0

so the values of x are (-5),3

Answer 2

$\large\underline{\bold{Given \:Question - }}$

$\sf \: Solve \: by \: factorisation \: method : \: {(x + 1)}^{2} - 16 = 0$

$\large\underline{\bold{Solution-}}$

Concept Used :-

Factorisation method :-

• In the factorisation method, we reduce any algebraic or quadratic equation into its simpler form, where the equations are represented as the product of factors instead of expanding the brackets.

• The factors of any equation can be an integer, a variable or an algebraic expression itself.

Identity used :-

$\boxed{ \bf{ {(x)}^{2} - {(y)}^{2} = (x + y)(x - y)}}$

Let's solve the problem now!!

Given that,

$\sf \: {(x + 1)}^{2} - 16 = 0$

$\sf \: {(x + 1)}^{2} - {(4)}^{2} = 0$

$\sf \: (x + 1 + 4)(x + 1 - 4) = 0$

$\sf \: (x + 5)(x - 3) = 0$

$\boxed{ \bf{ \therefore \: x \: = \: - \: 5 \: \: \: or \: \: \: x \: = \: 3 \: }}$

Alternative method :-

Splitting of middle terms :-

• In order to factorize  x² + bx + c we have to find numbers p and q such that p + q = b and pq = c.

• After finding p and q, we split the middle term in the quadratic as px + qx and get desired factors by grouping the terms.

Given that

$\sf \: {(x + 1)}^{2} - 16 = 0$

$\sf \: {x}^{2} + 1 + 2x - 16 = 0$

$\sf \: {x}^{2} + 2x - 15 = 0$

$\sf \: {x}^{2} + 5x - 3x - 15 = 0$

$\sf \: x(x + 5) - 3(x + 5) = 0$

$\sf \: (x + 5)(x - 3) = 0$

$\sf \: x + 5 \: = \: 0 \: \: \: \: \: or \: \: \: \: \: \: x - 3 = 0$

$\boxed{ \bf{ \therefore \: x \: = \: - \: 5 \: \: \: or \: \: \: x \: = \: 3 \: }}$

Additional Information :-

Nature of roots :-

Let us consider a quadratic equation ax² + bx + c = 0, then nature of roots of quadratic equation depends upon Discriminant (D) of the quadratic equation.

• If Discriminant, D > 0, then roots of the equation are real and unequal.

• If Discriminant, D = 0, then roots of the equation are real and equal.

• If Discriminant, D < 0, then roots of the equation are unreal or complex or imaginary.

Where,

• Discriminant, D = b² - 4ac

Let us consider a quadratic equation ax² + bx + c = 0, then other method to solve this quadratic equation is Quadratic Formula, given by

$\sf \: x \: = \: \dfrac{ - b \: \pm \: \sqrt{ {b}^{2} - 4ac} }{2a}$