Question

let the resistance of an electrical component remains constant while the potential difference across the two and of the component decrease to half of its formal value but change will occur in the current through it ​

Answer 1

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$\red{\bold{\underline{\underline{❥Question᎓}}}}$To factorise m^2-4m+33

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let the resistance of an electrical component remains constant while the potential difference across the two and of the component decrease to half of its formal value but change will occur in the current through it

${m}^{2} - 4 m + 33$

Here we use quadratic formula to find out the roots:-

Here ,a=1 ,b=-4 & c=33

$⟹m = \frac{ - b ± \sqrt{ {b}^{2} - 4ac} }{2a}$

$⟹ m= \frac{ - ( - 4)±\sqrt{ {( - 4)}^{2} - 4(1)(33)} }{2(1)}$

$⟹m= \frac{4± \sqrt{16 - 132} }{2}$

$⟹m = \frac{4± \sqrt{ - 116} }{2} = \frac{4±2 \sqrt{ - 29} }{2}$

$⟹m = \frac{2(2± \sqrt{ - 29} )}{2} = 2± \sqrt{29i}$

Here is your answer

i(we say iota is placed of -ve sign whenever inside the root value so we simply replace negative sign by placing iota instead of negative sign inside roots)

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Answer 2

Explanation:

When the potential difference across the two ends of the electrical component becomes half its former value, the current through it also becomes half. Since I = V/R, when potential difference becomes V/2, current becomes I/2 as the resistance (R) of the component remain' constant