Question

find a quadratic polynomial each with the given number as the sum and the product of it's zeros respectively (i) 4 and -2 (ii) 0 and -10/3 (iii) 5/7 and 0 (iv) -5 and -6 (v) √2 and -12 (vi) 3 and -2 (vii) -2√3 and -9 (viii) -3/2√5 and 1/2​

Answer 1

Given

we have given sum and Product of a quadratic polynomial:(i) 4 and -2 (ii) 0 and -10/3 (iii) 5/7 and 0 (iv) -5 and -6 (v) √2 and -12 (vi) 3 and -2 (vii) -2√3 and -9 (viii) -3/2√5 and 1/2

To Find

we have to find the quadratic polynomial

$\sf\huge {\underline{\underline{{Solution}}}}$

Since, we have given sum and Product of roots of the quadratic polynomial so, we can use this formula to find out the Equation.

x²-(sum of roots)x+ (product of roots)=0

(I) 4 and -2

x²-(4)x+(-2)=0

=> x²-4x-2=0

(ii) 0 and -10/3

x²-(0)x+(-10/3)=0

=> x²-10/3=0

(iii) 5/7 and 0

x²-(5/7)x+(0) =0

=> x²-5/7x=0

(iv) -5 and -6

x²-(-5)x+(-6)=0

=> x²+5x-6=0

(v) √2 and -12

x²-(√2)x+(-12)=0

=>x²-√2x-12=0

(vi) 3 and -2

x²-(3)x+(-2)=0

=>x²-3x-2=0

(vii) -2√3 and -9

x²-(-2√3)x+(-9)=0

=>x²+2√3x-9=0

(viii)-3/2√5 and 1/2

x²-(-3/2√5)x+(1/2)=0

=>x²+3/2√5x+1/2=0

Check:

we can check by factorise the obtained polynomial

(iv) -5 and -6

x²-(-5)x+(-6)=0

=> x²+5x-6=0

Now factorise it

x= -b±√ b²-4ac/2a

a= 1 ,b= 5 & c = -6

x= -5±√5²-4(1)(-6)/2

x= -5±√ 25-(-24)/2

x= -5±√ 25+24/2

x= -5±√49/2

x= -5±7/2

x= -5+7/2 or x = -5-7/2

x= 2/2 or x= -12/2

x= 1 or x = -6

sum of zeroes = -6+1= - 5

product of zeroes= -6*1= -6