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Question

(D) The given quadratic equation is $\sqrt{2}x^2 + 7x + 5\sqrt{2} = 0$. Comparing this with the standard form $ax^2 + bx + c = 0$,we get $a = \sqrt{2}

Vedclass · Accepted Answer

$-\sqrt{2}, -\frac{5}{\sqrt{2}}$

Vedclass · Answer

(D) The given quadratic equation is $\sqrt{2}x^2 + 7x + 5\sqrt{2} = 0$. Comparing this with the standard form $ax^2 + bx + c = 0$,we get $a = \sqrt{2}$,$b = 7$,and $c = 5\sqrt{2}$. The discriminant $D$ is given by $D = b^2 - 4ac$. $D = (7)^2 - 4(\sqrt{2})(5\sqrt{2}) = 49 - 4(5)(2) = 49 - 40 = 9$. Since $D > 0$,the equation has two distinct real roots. The roots are given by the quadratic formula $x = \frac{-b \pm \sqrt{D}}{2a}$. $x = \frac{-7 \pm \sqrt{9}}{2\sqrt{2}} = \frac{-7 \pm 3}{2\sqrt{2}}$. Case $1$: $x = \frac{-7 + 3}{2\sqrt{2}} = \frac{-4}{2\sqrt{2}} = -\frac{2}{\sqrt{2}} = -\sqrt{2}$. Case $2$: $x = \frac{-7 - 3}{2\sqrt{2}} = \frac{-10}{2\sqrt{2}} = -\frac{5}{\sqrt{2}}$. Thus,the solutions are $-\sqrt{2}$ and $-\frac{5}{\sqrt{2}}$.

Solve the following equation using the quadratic formula,if the equation has a solution in $R$: $\sqrt{2} x^{2} + 7x + 5\sqrt{2} = 0$

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