Question

If the roots of the equations \[\left( a^2 + b^2 \right) x^2 - 2b\left( a + c \right)x + \left( b^2 + c^2 \right) = 0\] are equal, then

पर्याय

• 2b = a + c

• b² = ac

• \[b = \frac{2ac}{a + c}\]

• b = ac

Answer 1

The given quadric equation is  \[\left( a^2 + b^2 \right) x^2 - 2b\left( a + c \right)x + \left( b^2 + c^2 \right) = 0\], and roots are equal.

Here,  `a = (a^2 +b^2), b = -2b(a+c) and, c = b^2 +c^2`

As we know that `D = b^2 - 4ac`

Putting the value of  `a = (a^2 +b^2), b = -2b (a + c) and, c = b^2 + c^2`

`= {-2b (a+c)}^2 - 4 xx (a^2 + b^2) xx (b^2 + c^2)`

`= 4a^2b^2 + 4b^2 c^2 + 8ab^2c - 4(a^2 b^2 + a^2 c^2 + b^4 + b^2 c^2)`

`=4a^2b^2 + 4b^2c^2 + 8ab^2c - 4a^2 b^2 - 4a^2c^2 - 4b^4 - 4b^2c^2`

`= +8ab^2c -4a^2c^2 - 4b^4`

`-4(a^2c^2 +b^4 - 2ab^2c)`

The given equation will have equal roots, if D =0

`-4(a^2c^2 +b^4 - 2ab^2c) = 0`

          `a^2c^2 +b^4 - 2ab^2 c = 0`

                       `(ac - b^2)^2 = 0`

                            `ac  -b^2 = 0`

                                       `ac = b^2`