Question

solve the following quadratic equations by formula method
$\sqrt{2 {x}^{2} } - 6x \times 3 \sqrt{2 = 0$
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Answer 1

Correct Question :

Q. Solve the following quadratic equation by formula method

$\tt \sqrt{2} {x}^{2} - 6x + 3 \sqrt{2} = 0.$

AnsWer :

x = 3 + √3/√2 and x = 3 - √3 /√2.

Solution :

We have,

Quadratic Equation,

$\tt \dagger \: \: \: \: \: \sqrt{2} {x}^{2} - 6x + 3 \sqrt{2} = 0.$

We have, two method to Find its zeros.

• Splitting the Middle term.
• Quadratic Formula.

Compare with General Equation.

$\tt \dagger \: \: \: \: \: a {x}^{2} + bx + c = 0. \: \: \: \: \: \red{a \neq0}.$

Let try To solve by Quadratic Formula.

$\tt \dagger \: \: \: \: \: x = \dfrac{ - b \pm \sqrt{ {b}^{2} - 4ac } }{2a}$

Where as,

• a = √2.
• b = -6.
• c = 3√2.

$\tt\mapsto x = \dfrac{6 \pm \sqrt{ { (- 6)}^{2} - 4 \times \sqrt{2} \times 3 \sqrt{2} } }{2 \sqrt{2} }$

$\tt\mapsto x = \dfrac{6 \pm \sqrt{36 - 24} }{2 \sqrt{2} }$

$\tt\mapsto x = \dfrac{6 \pm \sqrt{12} }{2 \sqrt{2} }$

$\tt\mapsto x = \dfrac{6 \pm2 \sqrt{3} }{2 \sqrt{2} }$

$\tt\mapsto x = \dfrac{2(3 \pm \sqrt{3}) }{2 \sqrt{2} }$

$\tt\mapsto x = \dfrac{3 \pm \sqrt{3} }{ \sqrt{2} }$

Either,

$\tt x = \dfrac{3 + \sqrt{3} }{ \sqrt{2} } \: \: \: \red{ or} \: \: \: x = \dfrac{3 - \sqrt{3} }{ \sqrt{2} }$