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2 years ago

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Solve x2-14x+21=0 by completing the square

1 answer:

ankoles [38]2 years ago

4 0

$x^2-14x+21=0\\\\x^2-2x\cdot7+7^2-7^2+21=0\\\\(x-7)^2-49+21=0\\\\(x-7)^2-28=0\\\\(x-7)^2=28\iff x-7=-\sqrt{28}\ \vee\ x-7=\sqrt{28}\\\\x=7-\sqrt{4\cdot7}\ \vee\ x=7+\sqrt{4\cdot7}\\\\x=7-2\sqrt7\ \vee\ x=7+2\sqrt7$

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<u>Step-by-step explanation:</u>

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Step-by-step explanation:

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antiseptic1488 [7]

Answer:

$\sf -11+7\sqrt{2}$

Step-by-step explanation:

Given expression:

$\sf \dfrac{3-\sqrt{32}}{1+\sqrt{2} }$

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$\sf \implies \dfrac{3-\sqrt{16 \cdot 2}}{1+\sqrt{2} }$

Apply radical rule $\sf \sqrt{a \cdot b}=\sqrt{a}\sqrt{b}$

$\sf \implies \dfrac{3-\sqrt{16}\sqrt{2}}{1+\sqrt{2} }$

As $\sf \sqrt{16}=4$ :

$\sf \implies \dfrac{3-4\sqrt{2}}{1+\sqrt{2} }$

Multiply by the conjugate:

$\sf \implies \dfrac{3-4\sqrt{2}}{1+\sqrt{2} } \times \dfrac{1-\sqrt{2} }{1-\sqrt{2} }$

$\sf \implies \dfrac{(3-4\sqrt{2})(1-\sqrt{2})}{(1+\sqrt{2})(1-\sqrt{2})}$

$\sf \implies \dfrac{3-3\sqrt{2}-4\sqrt{2}+4\sqrt{2}\sqrt{2}}{1-\sqrt{2}+\sqrt{2}-\sqrt{2}\sqrt{2}}$

As $\sf \sqrt{2}\sqrt{2}=\sqrt{4}=2$ :

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