Question

slove the quadratic equation : x sq -8x+12=0​

Answer 1

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Answer 2

To solve:

• x²-8x+12=0

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Solution:

~By middle term splitting

» x²-8x+12=0

» x²-6x-2x+12=0

» x(x-6)-2(x-6)=0

» (x-2)(x-6)=0

» x=2,6

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~By quadratic formula

• We have formula for finding roots::

$\underline{\boxed{\bf x= \dfrac{-b\pm\sqrt {b^2-4ac}}{2a}}}$

$\sf Here \begin{cases}\sf x=roots\;of\; equation\\\sf b=coefficient\:of\:x\\\sf a=coefficient\:of\:x^2\\\sf c=constant\:term \end{cases}$

• Given equation:

x²-8x+12=0

$\sf Here\begin{cases}\sf Coefficient\:of\:x^2,a=1\\\sf Coefficient\:of\:x,b=-8\\\sf Constant \:term,c=12\end{cases}$

Put these values in formula::

» $\bf x= \dfrac{-b\pm\sqrt {b^2-4ac}}{2a}$

⇒ $\sf x= \dfrac{-(-8)\pm\sqrt {(-8)^2-4(1)(12)}}{2(1)}$

⇒ $\sf x= \dfrac{8\pm\sqrt {64-48}}{2}$

⇒ $\sf x= \dfrac{8\pm\sqrt {16}}{2}$

⇒ $\sf x= \dfrac{8\pm4}{2}$

⇒ $\sf x= \dfrac{8+4}{2},\dfrac{8-4}{2}$

⇒ $\sf x= \dfrac{12}{2},\dfrac{4}{2}$

⇒ $\large\underline{\boxed{\sf\star {\red{x= 6,2}}}}$

Hence these are the required roots.

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