Find the roots of the following quadratic equ | vedclass

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Question

(A) 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}, c

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) 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 $a = 2, b = -2\sqrt{2}, c = 1$.First,calculate the discriminant $D = b^2 - 4ac$.$D = (-2\sqrt{2})^2 - 4(2)(1) = 8 - 8 = 0$.Since $D = 0$,the equation has two equal real roots.The quadratic formula is $x = \frac{-b \pm \sqrt{D}}{2a}$.Substituting the values,$x = \frac{-(-2\sqrt{2}) \pm \sqrt{0}}{2(2)} = \frac{2\sqrt{2}}{4} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$.Thus,the roots are $\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}$.

Find the roots of the following quadratic equation,if they exist,using the quadratic formula: $2x^2 - 2\sqrt{2}x + 1 = 0$.

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