Question

6. The roots of quadratic equation x2-5x+6=0 are
_and _​

Answer 1

Answer:

x=3 and x=2

Step-by-step explanation:

your answer

Answer 2

Given:-

• p(x) = x² - 5x + 6 = 0

To Find:-

Zeroes of the equation.

Solution:-

$\sf{x^2 - 5x + 6 = 0}$

We know,

When an equation is in the form ax² + bx + c = 0, the formula used to find the zeroes of the equation is:-

$\sf{\dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}}$

Here in this equation:-

a = Coefficient of x² = 1

b = Coefficient of x = -5

c = Constant Term = 6

Putting the values in the formula,

$\sf{\dfrac{-(-5) \pm \sqrt{(-5)^2 - 4\times 1\times 6}}{2\times1}}$

= $\sf{\dfrac{5\pm \sqrt{25 - 24}}{2}}$

= $\sf{\dfrac{5\pm \sqrt{1}}{2}}$

= $\sf{\dfrac{5+1}{2}\:\:and\:\:\dfrac{5-1}{2}}$

= $\sf{\dfrac{6}{2}\:\:and\:\:\dfrac{4}{2}}$

= $\sf{3\:\:and\:\:2}$

Therefore the two zeroes of the equation x² - 5x + 6 are 3 and 2

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Verification!!

We know,

Sum of zeroes = $\sf{\dfrac{-Coefficient\:of\:x}{Coefficient\:of\:x^2}}$

= $\sf{3+2 = \dfrac{-(-5)}{1}}$

= $\sf{5 = 5}$

Also

Product of zeroes = ${\sf{\dfrac{Constant\:Term}{Coefficient\:of\:x^2}}}$

= $\sf{3\times 2 = \dfrac{6}{1}}$

= $\sf{6 = 6}$

Hence Verified!

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Formula:-

When an equation is in the form ax²+bx+c=0 the formula for finding it's zeroes is:$\sf{\dfrac{-b\pm \sqrt{b^2 - 4ac}}{2a}}$

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