Question

यदि ABC वक्र को x2-7x+12, द्वारा दर्शाया गया है तो इसके शून्यांक होंगे |
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Answer 1

Answer:

3 and 4

Step-by-step explanation:

x² - 7x + 12

take two numbers whose product is 12 and sum or difference is -7

-4 × -3 = 12

-4 + (-3) = -7

x² - 4x - 3x + 12

x ( x - 4) - 3 ( x - 4)

(x - 4)(x - 3)

If we make the polynomial an equation, so

x² - 7x + 12 = 0

(x - 4)(x - 3) = 0

since the product is zero, one of the factor is zero

if (x - 4) is 0

x - 3 = 0

x = 3

if (x - 3) is 0

x - 4 = 0

x = 4

so the zeroes of x² - 7x + 12 are 3 and 4

TIP : The number of roots of an equation is equal to the highest degree of the equation.

example : the highest degree if the equation is 2 (x²), so the number of zeroes of the equation is 2.

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Answer 2

Given :  ABC वक्र को x²-7x+ 12, द्वारा दर्शाया गया है

To Find : शून्यांक  

(i) (2,-3)

(ii)3,4)

(iii) (4,-5)

(IV)3,-5

Solution:

x²-7x+ 12,

= x² - 4x - 3x + 12

= x(x - 4) - 3(x - 4)

= (x - 3)(x - 4)

x - 3 = 0=> x = 3

x - 4 = 0 => x = 4

शून्यांक    3 , 4

Learn More:

2) शून्यक -2 और 5 वाले बहुपदों की ...

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