Question

Roots of the equation X2-14X+49=0 is one. is it real and equal, real and unequal​

Answer 1

Answer:

real and equal

Step-by-step explanation:

Comparing x^2 - 14x + 49 = 0 with ax^2 + bx + c = 0:

a = 1    ;  b = - 14   ; c = 49

     Discriminant = b^2 - 4ac

                     = (-14)² - 4(1)(49)

                     = 196 - 196

                     = 0

As the discriminant is 0, roots are real and equal.

Answer 2

• We have to find the root of this equation is real and equal , real and unequal ?

Now,

$:\implies\sf x^2-14x+49$

• After finding discriminate we can find the roots are real and equal or real and unequal

$:\implies\sf \sqrt{D}= b^2-4ac\\ \\ \\ \bullet\sf \ a= 1 ,\ \ b= -14\ \, c= 49\\ \\ \\ :\implies\sf \sqrt{D}= (-14)^2-4\times 49\times 1\\ \\ \\ :\implies\sf \sqrt{D}=196-196\\ \\ \\ :\implies\sf \sqrt{D}= 0$

So the roots are real and equal