How do you solve $\sqrt{4 z + 1} = 3 + \sqrt{4 z - 2}$ $sqrt(4z+1)=3+sqrt(4z-2)$ ?

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Answer 1 · 2015-04-14T15:10:03

$z = \frac{3}{4}$ $z=3/4$

Solution

$\sqrt{4 z + 1} = 3 + \sqrt{4 z - 2}$ $sqrt(4z+1)=3+sqrt(4z-2)$

Squaring both sides

${(\sqrt{4 z + 1})}^{2} = {(3 + \sqrt{4 z - 2})}^{2}$ $(sqrt(4z+1))^2=(3+sqrt(4z-2))^2$

$\Rightarrow (4 z + 1) = {(3)}^{2} + {(\sqrt{4 z - 2})}^{2} + 2 (3) \sqrt{4 z - 2}$ $=>(4z+1)=(3)^2+(sqrt(4z-2))^2+2(3)sqrt(4z-2)$

$\Rightarrow 4 z + 1 = 9 + 4 z - 2 + 6 \sqrt{4 z - 2}$ $=>4z+1=9+4z-2+6sqrt(4z-2)$

$\Rightarrow 0 = - 4 z - 1 + 9 + 4 z - 2 + 6 \sqrt{4 z - 2}$ $=>0=-4z-1+9+4z-2+6sqrt(4z-2)$

$=>0=-cancel(4z)-1+9+cancel(4z)-2+6sqrt(4z-2)$

$=>0=6+6sqrt(4z-2)$

$=>-6=6sqrt(4z-2)$

$=>-6/6=sqrt(4z-2)$

$=>-1=sqrt(4z-2)$

Again squaring both sides

$(-1)^2=(sqrt(4z-2))^2$

$=>1=4z-2$

$=>1+2=4z$

$=>3=4z$

$=>3/4=z$

Or

$z=3/4$

How do you solve sqrt(4z+1)=3+sqrt(4z-2)√4z+1=3+√4z−2sqrt(4z+1)=3+sqrt(4z-2)?