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Q 3.

Expert-verifiedFound in: Page 643

Book edition
6th

Author(s)
Sullivan

Pages
1200 pages

ISBN
9780321795465

Use the Square Root Method to find the real solutions of ${(x+4)}^{2}=9$.

Real solution of ${(x+4)}^{2}=9$ is x = -1 and $x=-7$.

We have given equation is,

${(x+4)}^{2}=9$

If ${a}^{2}=bthena=\sqrt{b}anda=-\sqrt{b}$.

We have given,

${(x+4)}^{2}=9$

Using square root method,

localid="1646649530505" $(x+4)=\sqrt{9}and(x+4)=-\sqrt{9}$

Solving both equations for x,

$(x+4)=3and(x+4)=-3\phantom{\rule{0ex}{0ex}}\Rightarrow x=3-4andx=-3-4\phantom{\rule{0ex}{0ex}}\Rightarrow x=-1andx=-7$

Hence, the real solution for ${(x+4)}^{2}=9$ is

$x=-1$ and $x=-7$

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