Question

a) Find the roots of quadratic equation 2x² - 2√2x + 1 = 0.​

Answer 1

Given:-

• Quadratic equation = 2x² - 2√2x + 1 = 0

To Find:-

• roots of the equation

Solution:-

We know, when an equation is in the form ax² + bx + c = 0, it is said to be quadratic equation, where:-

• a = Coefficient of x²
• b = Coefficient of x
• c = Constant term

Here,

2x² - 2√2x + 1 = 0 is in the form ax² + bx + c = 0, where:-

• a = 2
• b = -2√2
• c = 1

To find roots of a quadratic equation we use the formula:-

• $\sf{\dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}}$

Putting all the values in the formula:-

$= \sf{\dfrac{-(-2\sqrt{2}) \pm \sqrt{(-2\sqrt{2})^2 - 4\times 2 \times 1}}{2\times 2}}$

$= \sf{\dfrac{2\sqrt{2} \pm \sqrt{8 - 8}}{4}}$

$= \sf{\dfrac{2\sqrt{2} \pm \sqrt{0}}{4}}$

$= \sf{\dfrac{2\sqrt{2} + 0}{4}\:\:and\:\:\dfrac{2\sqrt{2} - 0}{4}}$

$= \sf{\dfrac{\not{2\sqrt{2}}}{\not{4}}\:\:and\:\:\dfrac{\not{2\sqrt{2}}}{\not{4}}}$

$= \sf{\dfrac{\sqrt{2}}{2}\:\:and\:\:\dfrac{\sqrt{2}}{2}}$

$\underline{\boxed{\pink{\rm{\therefore\:The\:two\:roots\:of\:the\:equation\:are\:\dfrac{\sqrt{2}}{2}\:and\:\dfrac{\sqrt{2}}{2}}}}}$

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Answer 2

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