Question

√2x + 4 = 1o-x (in icse class 10th )
chapter 5 quadratic equation (exercise 5.3)
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Answer 1

Answer:

√2x+x=10-4

(√2+1)x=6

x=6/(√2+1)

x=6*(√2-1)/2-1

x=6(√2-1)

Answer 2

$\sf{\underline{\boxed{\green{\large{\bold{ Solution}}}}}}$

$\sf\implies \sqrt 2 x + 4 = 10 - x$

$\sf\implies\sqrt 2 x = 10 - 4 - x$

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$\sf\implies\sqrt 2 x = 6 - x$ ⠀⠀⠀

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• Squaring both the sides

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$\sf\implies ( \sqrt 2 x)^2 = ( 6- x)^2$ ⠀⠀

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$\sf\implies 2x^2 = 36 + x^2 - 12x$

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$\sf\implies 2x^2- x^2+ 12x - 36= 0$

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$\sf\implies x^2 + 12x - 36 = 0$

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compare the eq with $\sf{\underline{\bold{ax^2 + bx + c = 0 }}}$

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☯ a = 1

☯ b = 12

☯ c = -36

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now :-

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$\sf{\underline{\boxed{\pink{\large{\mathfrak{x = \dfrac{ - b \pm \sqrt D }{2a }}}}}}}$

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$\sf{\underline{\boxed{\pink{\large{\mathfrak{ D = b^2 - 4ac }}}}}}$

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finding value of D.

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$\sf\implies D = b^2 - 4ac$

$\sf\implies D = (12)^2 - 4 \times 1 \times -36$

$\sf\implies D = 144 + 144$

$\sf\implies D = 288$

$\sf{\underline{\boxed{\blue{\large{\bold{ D = 288}}}}}}$

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putting values in the eq.

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$\sf\implies x = \dfrac{ -b \pm\sqrt D }{2a}$

$\sf\implies x = \dfrac{ -( 12 ) \pm\sqrt {288} }{2\times 1 }$

$\sf\implies x = \dfrac{ -12 \pm 12\sqrt 2}{2}$

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✒$\sf x = \dfrac{ -12 + 12\sqrt 2 }{ 2 }$

$\implies x = -6 + 6\sqrt 2$

$\sf{\underline{\boxed{\purple{\large{\bold{ x = -6 + 6\sqrt 2}}}}}}$

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✒$\sf x = \dfrac{ -12 - 12\sqrt 2 }{ 2 }$

$\implies x = -6 -6\sqrt2$

$\sf{\underline{\boxed{\purple{\large{\bold{ x = -6 -6\sqrt 2}}}}}}$

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$\sf{\underline{\boxed{\purple{\large{\bold{ x = -6 + 6\sqrt 2 \: or -6 - 6\sqrt 2 }}}}}}$

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