Obtain the roots of the following quadratic e | vedclass

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Question

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

Vedclass · Accepted Answer

$\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}$

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(B) The given quadratic equation is $2x^{2} - 2\sqrt{2}x + 1 = 0$.Comparing this with the standard form $ax^{2} + bx + c = 0$, we get $a = 2$, $b = -2\sqrt{2}$, and $c = 1$.The quadratic formula is $x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$.First, calculate the discriminant $D = b^{2} - 4ac = (-2\sqrt{2})^{2} - 4(2)(1) = 8 - 8 = 0$.Since $D = 0$, the equation has two equal real roots.The roots are $x = \frac{-(-2\sqrt{2}) \pm \sqrt{0}}{2(2)} = \frac{2\sqrt{2}}{4} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$.Thus, the roots are $\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}$.

Obtain the roots of the following quadratic equation by using the general formula for the solution: $2x^{2} - 2\sqrt{2}x + 1 = 0$.

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