Exercise ID206

Algebra → Quadratic equations, inequalities and systems of equations → Equations Reducible to Quadratic Form
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Solve the following equation.

$\sqrt{2x-3}=\sqrt{x+10}-1$

$\sqrt{2x-3}=\sqrt{x+10}-1\phantom{\rule{0ex}{0ex}}$ ${\left(\sqrt{2x-3}\right)}^{2}={\left(\sqrt{x+10}-1\right)}^{2}\phantom{\rule{0ex}{0ex}}$ $2x-3=x+10-2\sqrt{x+10}+1\phantom{\rule{0ex}{0ex}}$ $2\sqrt{x+10}=x+10+1-2x+3\phantom{\rule{0ex}{0ex}}$ ${\left(2\sqrt{x+10}\right)}^{2}={\left(14-x\right)}^{2}\phantom{\rule{0ex}{0ex}}$ $4\left(x+10\right)=196-28x+{x}^{2}\phantom{\rule{0ex}{0ex}}$ $4x+40=196-28x+{x}^{2}\phantom{\rule{0ex}{0ex}}$ $196-28x+{x}^{2}-4x-40=0\phantom{\rule{0ex}{0ex}}$ ${x}^{2}-32x+156=0\phantom{\rule{0ex}{0ex}}$ ${x}_{1,2}=\frac{32±\sqrt{{\left(-32\right)}^{2}-4·1·156}}{2}=\frac{32±\sqrt{400}}{2}=\frac{32±20}{2}\phantom{\rule{0ex}{0ex}}$ $x=6\phantom{\rule{0ex}{0ex}}$
$x=6$

${x}_{1,2}=\frac{-b±\sqrt{{b}^{2}-4ac}}{2a};a{x}^{2}+bx+c=0$