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Question

(A) The given quadratic equation is $x^{2} + 6x + 5 = 0$.Comparing this with the standard form $ax^{2} + bx + c = 0$,we get $a = 1$,$b = 6$,and $c = 5

Vedclass · Accepted Answer

$-5, -1$

Vedclass · Answer

(A) The given quadratic equation is $x^{2} + 6x + 5 = 0$.Comparing this with the standard form $ax^{2} + bx + c = 0$,we get $a = 1$,$b = 6$,and $c = 5$.The quadratic formula is $x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$.First,calculate the discriminant $D = b^{2} - 4ac = (6)^{2} - 4(1)(5) = 36 - 20 = 16$.Since $D > 0$,real solutions exist.Substituting the values into the formula: $x = \frac{-6 \pm \sqrt{16}}{2(1)} = \frac{-6 \pm 4}{2}$.For the positive sign: $x = \frac{-6 + 4}{2} = \frac{-2}{2} = -1$.For the negative sign: $x = \frac{-6 - 4}{2} = \frac{-10}{2} = -5$.Thus,the solutions are $-5$ and $-1$.

Solve the following equation using the quadratic formula,if the equation has a solution in $R$: $x^{2}+6x+5=0$.

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