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
1 thought on “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 a”
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
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
[tex]\sf\huge {\underline{\underline{{Solution}}}}[/tex]
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