Question

solve quadratic equation by factorisation a.
$10 {x}^{2} - 20x + 10 = 0$
b.
${x}^{2} - 4x + 1 = 0$
​

Answer 1

Answer:

a. 10x² - 20x + 10 = 0

» 10x² - 10x - 10x + 10 = 0

» 10x (x - 1) - 10 (x - 1) = 0

» (10x - 10) (x - 1) = 0

» x = 1 or 1

b. x² - 4x + 1 = 0

» x = [-b ± √(b² - 4ac)]/2a

» x = [4 ± √(16 - 4)]/2

» x = [4 ± 2√3]/2

» x = 2 ± √3

Answer 2

Step-by-step explanation:

Given:-

The equations are

1)10x^2-20x+10=0

2)x^2-4x+1=0

To find:-

solve the quadratic equations by factorisation method.

Solution:-

a)

Given equation is 10x^2-20x+10=0

=>10x^2 -10x -10x +10 = 0

=>10x(x-1) -10(x-1) = 0

=>(x-1)(10x-10)=0

=>10(x-1)(x-1) = 0

=>(x-1)(x-1) = 0

=>x-1 = 0 or x-1 = 0

=>x = 1 and 1

(or)

Given equation is 10x^2-20x+10=0

=>10(x^2-2x+1)=0

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

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

It is in the form a^2-2ab+b^2

we know that a^2-2ab+b^2 = (a-b)^2

=>(x-1)^2 = 0

=>(x-)(x-1) = 0

=>x-1 = 0 or x-1 = 0

=>x= 1 and 1

Answer:-

Solution for the given equation is (1,1)

b) Given equation is x^2 -4x +1 = 0

=>x^2 -4x +1 = 0

It can be written as

=>x^2 - 4x + 4 - 3 = 0

=>(x^2-4x+4) -3 = 0

=>(x^2-2(x)(2)+2^2) -3 = 0

=>(x-2)^2 -3 = 0

=>(x-2)^2 -(√3)^2 = 0

It is in the form of a^2-b^2

where a= (x-2)

b= √3

we know that

a^2-b^2 = (a+b)(a-b)

=>(x-2+√3)(x-2-√3) = 0

=>x-2+√3 = 0 or x-2-√3 = 0

=>x= 2-√3 or x= 2+√3

Answer:-

Solution for the given equation is 2-√3 and 2+√3

Used formulae:-

• a^2-b^2 = (a+b)(a-b)
• (a-b)^2 = a^2-2ab+b^2