Question

Find the zeroes of quadratic polynomial x^2 -5 .also verify relation btw zeroes and coefficients

Answer 1

Answer:

hope this helps u

Step-by-step explanation:

x^2= 5

x= √5

x= ± √5

we consider alpha(α)= √5

                       beta(β)=  - √5

alpha + beta= - b/a

√5 +(-√5)= -0/1

0=0

alphabeta= c/a

√5*-√5= -5/1

-5= -5

Answer 2

Answer:

${x}^{2} - 5$

• LET P(X)=0
• ${x}^{2} - 5 = 0$
• ${x}^{2} = 5$
• $x = \sqrt[ + - ]{5}$
• $\alpha = \sqrt{5}$
• $\beta = - \sqrt{5}$
• ${x}^{2} co.efficient = 1$
• X Co-efficient =o
• constant = C = -5
• SUM OF ZEROES =alpha+beta= -b/a
• 0/1=0.
• $\alpha + \beta = - \sqrt{5} + \sqrt{5} = 0$
• PRODUCT OF ZEROES =alpha(beta)=c/a
• $\alpha \times \beta = - \sqrt{5} \times \sqrt{5} = - 5$
• C/A= -5/1= -5

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