Question

Show that: ${(3\mathrm{x}+7)}^2-84\mathrm{x}={(3\mathrm{x}-7)}^2$

Answer 1

To prove :${(3\mathrm{x}+7)}^2-84\mathrm{x}={(3\mathrm{x}-7)}^2$

$\mathrm{L}.\mathrm{H}.\mathrm{S}$ $={(3\mathrm{x}+7)}^2-84\mathrm{x}$

$={\left(3\mathrm{x}\right)}^2+7^2+2\times 3\mathrm{x}\times 7-84\mathrm{x}$ [ Formula used : ${(\mathrm{a}+\mathrm{b})}^2={\mathrm{a}}^2+{\mathrm{b}}^2+2\mathrm{ab}$]

$={9\mathrm{x}}^2+49+42\mathrm{x}-84\mathrm{x}$

$={9\mathrm{x}}^2-42\mathrm{x}+49$

$\mathrm{R}.\mathrm{H}.\mathrm{S}$ $={(3\mathrm{x}+7)}^2$

$={\left(3\mathrm{x}\right)}^2+7^2+2\times 3\mathrm{x}\times 7$

$=9{\mathrm{x}}^2+49-42\mathrm{x}$ [ Formula used : ${(\mathrm{a}+\mathrm{b})}^2={\mathrm{a}}^2+{\mathrm{b}}^2-2\mathrm{ab}$]

$=9{\mathrm{x}}^2-42\mathrm{x}+49$

From above it is clear that $\mathrm{L}.\mathrm{H}.\mathrm{S}$ $=$ $\mathrm{R}.\mathrm{H}.\mathrm{S}$

Thus, ${(3\mathrm{x}+7)}^2-84\mathrm{x}={(3\mathrm{x}-7)}^2$ .