Question

x square - 2 X + a square minus b square equals to zero by quadratic formula​

Answer 1

Answer:

x^2-2ax+(a^2-b^2) =0

quadratic formula

X= (-b+ or- √(b^2-4ac) )/ 2a

here a=1, b=-2a, c= a^2-b^2

X= ( 2a + or - √( 4a^2-4a^2+4b^2))/2

X= ( 2a + or - √4b^2)/2

X= (2a+2b)/2 , ( 2a-2b)/2

X= a+b , a-b

Answer 2

Correct question

x square - 2 X + a square minus b square equals to zero by quadratic formula​.Find the roots of the equation.

Answer:

The roots of the equation are  $x=-1+\sqrt{1-(a^{2} -b^{2}) }$ and$x=-1-\sqrt{1-(a^{2} -b^{2}) }$.

Step-by-step explanation:

Given:

Quadratic equation $x^{2} -2x+a^{2} -b^{2} =0$

To Find:

To find the roots of equation.

Formula Used:

If $ax^{2} +bx+c=0$ is quadratic equation.

Therefore the roots of the given equation can be found by

$x=\frac{-b\±\sqrt{b^{2}-4ac } }{2a}$   --------- formula no. 01.

Solution:

As given-Quadratic equation $x^{2} -2x+a^{2} -b^{2} =0$

Comparing quadratic equation with $ax^{2} +bx+c=0$.

a=1, b= -2 and $c=a^{2} -b^{2}$

Applying the formula no. 01.

$x=\frac{-(-2)\±\sqrt{(-2)^{2}-4\times 1\times (a^{2} -b^{2} ) } }{2\times1}$

    $=\frac{2\±\sqrt{4-4 (a^{2} -b^{2} ) } }{2}$

   $=\frac{2\±\sqrt{4[(1- (a^{2} -b^{2} )] } }{2}$

   $=\frac{2\±2\sqrt{[(1- (a^{2} -b^{2} )] } }{2}$

  $x= 1 \± \sqrt{[(1- (a^{2} -b^{2} )]$

$x=-1+\sqrt{1-(a^{2} -b^{2}) }$ and$x=-1-\sqrt{1-(a^{2} -b^{2}) }$.

Thus, the  roots of the equation are  $x=-1+\sqrt{1-(a^{2} -b^{2}) }$ and$x=-1-\sqrt{1-(a^{2} -b^{2}) }$.

PROJECT CODE #SPJ3