Question

Find the roots of the equation, if they exists, by applying the quadratic formula :

$4x {}^{2} + 4bx - (a {}^{2} - b {}^{2} ) = 0$

Answer 1

Given Equation is 4x^2 + 4bx - (a^2 - b^2) = 0.

Here, a = 4, b = 4b, c = -(a^2 - b^2) = -a^2 + b^2

(i)

$=> x=\frac{-b+\sqrt{b^2-4ac}}{2a}$

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

$=>\frac{-4b+\sqrt{16b^2+16a^2-16b^2}}{8}$

$=> \frac{-4b+ \sqrt{16a^2}}{8}$

$=> \frac{-4b+4a}{8}$

$=> \frac{4(b-a)}{8}$

$=> \frac{b-a}{2}$

(ii)

$=>x =\frac{-b-\sqrt{b^2-4ac}}{2a}$

$=>\frac{-4b- \sqrt{(4b)^2 - 4(4)(-a^2 + b^2}}{2*4}$

$=>\frac{-4b- \sqrt{16b^2 + 16a^2 - 16b^2}}{8}$

$=>\frac{-4b+ \sqrt{16a^2}}{8}$

$=>\frac{-4b-4a}{8}$

$=>-\frac{b+a}{2}$

Therefore, the roots of the equation are:

$=>x=\boxed{\frac{a-b}{2},-\frac{b+a}{2}}$

Hope it helps!

Answer 2

$\huge \bf \purple{Hey \: there !! }$


The given quadratic equation :-

$4x {}^{2} + 4bx - (a {}^{2} - b {}^{2} ) = 0$


Comparing it with Ax² + Bx + C = 0 , we get

A = 4, B = 4b and C = -( a² - b² ) .


•°• D = B² - 4AC .

= (4b)² - 4 × 4 × [ - ( a² - b² ) ] .

= 16b² + 16a² - 16b² .

= 16a² > 0.

So, the given equation has real roots .

$\alpha = \frac{ - B + \sqrt{D} }{2A} \\ \\ = \frac{ - 4b + \sqrt{16 {a}^{2} } }{2 \times 4} . \\ \\ = \frac{ - 4b + 4a}{8} . \\ \\ = \frac{4( - b +a) }{8} = \boxed{ \green{ \frac{1}{2} (a + b).}} \\ \\ and \\ \\ \beta = \frac{ - B - \sqrt{D} }{2A} \\ \\ = \frac{ - 4b - \sqrt{16 {a}^{2} } }{2 \times 4} \\ \\ = \frac{ - 4b - 4a}{8} . \\ \\ = \frac{ - 4(a + b)}{8} = \boxed{ \green{ \frac{ - 1}{2} (a + b).}}$


✔✔ Hence, it is solved ✅✅.



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