Question

The roots of x2-7x+12=0​

Answer 1

Answer:

${x}^{2} - 7x + 12 = {x}^{2} - 4x - 3x + 12 = x(x - 4) - 3(x - 4) = (x - 3)(x - 4)$

SO ROOTS ARE 3,4

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Answer 2

$\displaystyle\sf{Given,$

$\displaystyle\sf{x^2-7x+12=0$

$\displaystyle\sf{Here,x\:can\:be\:determined\:by\:the\:equation:$

$\boxed{\displaystyle\sf{x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}}}$

$\displaystyle\sf{Here,$

• $\displaystyle\sf{a=1$

• $\displaystyle\sf{b=-7$

• $\displaystyle\sf{c=12$

$\displaystyle\sf{Substituting\:the\:values\:to\:the\:equation,$

$\displaystyle\sf{\implies x=\frac{7\pm\sqrt{49-(4\times1\times12)}}{(2\times1)}$

$\displaystyle\sf{\implies x=\frac{7\pm\sqrt{49-48}}{2}$

$\displaystyle\sf{\implies x=\frac{7\pm 1}{2}$

$\displaystyle\sf{\implies x=\frac{7+1}{2}\:\:\:\:\:\:\:\:\displaystyle\sf{\implies x=\frac{7-1}{2}$

$\displaystyle\sf{\implies x=\frac{8}{2}\:\:\:\:\:\:\:\:\displaystyle\sf{\implies x=\frac{6}{2}$

$\displaystyle\sf{\implies x=4\:\:\:\:\:\:\:\:\displaystyle\sf{\implies x=3$

$\boxed{\displaystyle\sf{\therefore x\:can\:have\:the\:values=4\:,\:3}}$