Find the roots of the following quadratic equ | vedclass

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Question

(A) Given equation: $4x^{2} + 4bx - (a^{2} - b^{2}) = 0$.Divide by $4$: $x^{2} + bx - \frac{a^{2}-b^{2}}{4} = 0$.Add and subtract $(\frac{b}{2})^{2}$

Vedclass · Accepted Answer

$-\left(\frac{a+b}{2}\right), \left(\frac{a-b}{2}\right)$

Vedclass · Answer

(A) Given equation: $4x^{2} + 4bx - (a^{2} - b^{2}) = 0$.Divide by $4$: $x^{2} + bx - \frac{a^{2}-b^{2}}{4} = 0$.Add and subtract $(\frac{b}{2})^{2}$ to complete the square: $x^{2} + bx + (\frac{b}{2})^{2} - (\frac{b}{2})^{2} - \frac{a^{2}-b^{2}}{4} = 0$.$(x + \frac{b}{2})^{2} = \frac{b^{2}}{4} + \frac{a^{2}-b^{2}}{4} = \frac{a^{2}}{4}$.Taking square root on both sides: $x + \frac{b}{2} = \pm \frac{a}{2}$.Case $1$: $x = \frac{a-b}{2}$.Case $2$: $x = \frac{-a-b}{2} = -(\frac{a+b}{2})$.Thus,the roots are $-\left(\frac{a+b}{2}\right), \left(\frac{a-b}{2}\right)$.

Part $I$	Part $II$
$1.$ The discriminant of $x^{2}+5x+6=0$	$a. 1$
$2.$ The discriminant of $x^{2}+5x+4=0$	$b. 9$
$3.$ The discriminant of $x^{2}+4x+3=0$	$c. 4$
$4.$ The discriminant of $x^{2}+6x+5=0$	$d. 16$

Find the roots of the following quadratic equation by the method of completing the square: $4x^{2} + 4bx - (a^{2} - b^{2}) = 0$.

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