Question

the zeroes of the quadratic polynomial 4x^2+3x+2 is​

Answer 1

Given:

• Quadratic polynomial is 4x² + 3x + 2 =  0

To Find:

• The roots of the given polynomial

Basic Knowledge:

• A quadratic equation is in the form ax² + bx + c ,  where a ≠ 0
• There can be atmost 2 roots for a quadratic equation
• The roots of the equation are denoted by the symbols α and β

If alpha ( α ) is the root of the equation then the equation would be such that aα² + bα + c and in the same way if beta ( β ) is said to be thee root the equation would be aβ² + bβ + c

Solution:

Now here we observe that we have been provided with a polynomial of degree 2 and said to find thee roots of it so , now let's use quadratic formula to find the roots of the polynomial

Quadratic Formula :

$\bullet {\underline{\boxed{\tt{ x = \dfrac{- b \pm \sqrt{\triangle} }{2a} }}}}$        

Were the sign Δ is called delta or discriminant which stands for thee value b² - 4ac so, the quadratic formula can be written as :

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

So, now let's compare the constant terms and find the roots of the equation

→ a = 4

→ b = 3

→ c = 2

Now let's put the values in the formula and find the roots

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

$\longrightarrow \tt \alpha = \dfrac{- 3+\sqrt{3^2 - 4(4)(2)} }{2(4)}$

$\longrightarrow \tt \alpha = \dfrac{ -3 + \sqrt{9 - 32}}{8}$

$\longrightarrow \tt \alpha = \dfrac{ -3 + \sqrt{-23}}{8}$

Since delta < 0 and the root of a negetive integer doesn't exit so, the roots would be complex

$\longrightarrow \tt \alpha = \dfrac{ -3 + \sqrt{ 23}}{8} \; \sf i$

we know that the roots are conjugate to each other when they are complex

so the other root will be

$\longrightarrow \tt \beta = \dfrac{ -3 - \sqrt{ 23}}{8} \; \sf i$

Therefore:

• The roots are - 3 - √ 23 / 8 i and  - 3 + √ 23 / 8 i

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Answer 2

Answer:

same thing I can see it again and again and again