📺 Subscribe Our YouTube Channels: Doubtify JEE | Doubtify Class 10

Search Suggest

Form Quadratic Polynomial from Sum and Product of Zeroes

Learn how to form quadratic polynomials using the sum and product of zeroes with the formula:

 

❓ Question

Find a quadratic polynomial for each pair where the given numbers are respectively the:

  • Sum of zeroes (S)
  • Product of zeroes (P)
(i) 1/4, -1
(ii) √2, 1/3
(iii) 0, √5
(iv) 1, 1
(v) -1/4, 1/4
(vi) 4, 1

đź–Ľ️ Solution Image

Form Quadratic Polynomial from Sum and Product of Zeroes


✍️ Short Concept

For a quadratic polynomial:

ax2+bx+cax^2+bx+c

if:

Sum of zeroes=S\text{Sum of zeroes}=S

and

Product of zeroes=P\text{Product of zeroes}=P

then the polynomial is:

f(x)=x2Sx+Pf(x)=x^2-Sx+P

đź’Ż

If fractions are present, we multiply throughout to remove denominators.


đź”· (i) S=14, P=1S=\frac14,\ P=-1

Using:

f(x)=x214x1f(x)=x^2-\frac14x-1

Multiplying by 4:

4x2x4\boxed{4x^2-x-4}

đź”· (ii) S=2, P=13S=\sqrt2,\ P=\frac13

f(x)=x22x+13f(x)=x^2-\sqrt2x+\frac13

Multiplying by 3:

3x232x+1\boxed{3x^2-3\sqrt2x+1}

đź”· (iii) S=0, P=5S=0,\ P=\sqrt5

f(x)=x20x+5f(x)=x^2-0x+\sqrt5
x2+5\boxed{x^2+\sqrt5}

đź”· (iv) S=1, P=1S=1,\ P=1

f(x)=x2x+1f(x)=x^2-x+1
x2x+1\boxed{x^2-x+1}

đź”· (v) S=14, P=14S=-\frac14,\ P=\frac14

f(x)=x2+14x+14f(x)=x^2+\frac14x+\frac14

Multiplying by 4:

4x2+x+1\boxed{4x^2+x+1}

đź”· (vi) S=4, P=1S=4,\ P=1

f(x)=x24x+1f(x)=x^2-4x+1
x24x+1\boxed{x^2-4x+1}

✅ Final Answers

PartQuadratic Polynomial
(i)4x2x44x^2-x-4
(ii)3x232x+13x^2-3\sqrt2x+1
(iii)x2+5x^2+\sqrt5
(iv)x2x+1x^2-x+1
(v)4x2+x+14x^2+x+1
(vi)x24x+1x^2-4x+1

⭐ Important Formula

If zeroes are:

α,β\alpha,\beta

then:

α+β=S\alpha+\beta=S
αβ=P\alpha\beta=P

Required quadratic polynomial:

x2Sx+P\boxed{x^2-Sx+P}


📚 Related Topics

Post a Comment

Have a doubt? Drop it below and we'll help you out!