Question

X^2+2x-400=0

Use the factorisation method

Answer 1

Given,

The quadratic equation is x^+2x-400.

Now to solve a quadratic equation we can use 3 methods =>

1)Completing Square Method,

2)Middle term Split method, and

3)By Quadratic formula.

Here we will be using two methords =>the quadratic formula method and Middle term split methord to solve the quadratic equation as it is not suitable here to solve the equation by factorization method --->

1)Quadratic Formula Method

d=b^2 - 4×a ×c

=>d = (2)^2 -{4 × 1×(-400)}

=>d = 4 + 1600

=>d = 1604.

Hence we have =>

x = $\dfrac{-b\pm\sqrt d}{2\times{a}}$

=>x=$\dfrac{-2\pm\sqrt1604}{2\times1}$

=>x = $\dfrac{-2\pm\sqrt1604}{2}$

So,

either,

x= $\dfrac{-2+\sqrt1604}{2}$

=>x=$\dfrac{-2+2\sqrt401}{2}$

=>x=${-1+\sqrt401}$

or

x = $\dfrac{-2-\sqrt1604}{2}$

=>x=$\dfrac{-2-2\sqrt401}{2}$

=>x=${-1-\sqrt401}$

2)Middle term Split methord

$x^2$ +2x-400=0

=> $x^2$ + 2x + 1 - 400 - 1 = 0

=> $(x + 1)^2$ - 401 = 0

=>$(x + 1)^{2}$ - ${\sqrt{401}} ^2$ = 0

Now, by using ${a}^{2} - {b}^{2}$

= (a+b) (a-b) we get=>

=> (x + 1+ $\sqrt401$) (x + 1 - $\sqrt401$) = 0

So

either,

x=${-1+\sqrt401}$

Or

x=${-1-\sqrt401}$

REMEMBER

1)Here

d = discriminant (decides the nature of roots or zeroes of a quadratic equation.

a)when d = 0

then the quadratic equation will have the same real and equal roots.

b)when d>0

then the quadratic equation will have two real and distinct roots.

c)when d<0

then the quadratic equation will have no real roots i.e it will have imaginary roots.

Remember to use these methods according to the question given.