Question

Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.
(i) $\frac{1}{4}$, $-1$ (ii) $\sqrt{2},\frac{1}{3}$ (iii) $0,\sqrt{5}$
(iv) $1,1$ (v) $\frac{-1}{4},\frac{1}{4}$ (vi) $4,1$

Answer 1

(i) $\frac{1}{4}$, -1

Using the quadratic equation formula,
$x^2-\left(Sumofroot\right)x+\left(Productofroot\right)=0$
Substitute the value in the formula, we get
$x^2-\frac{1}{4}x-1=0$
$4x^2-x-4=0$

(ii) $\sqrt{2},\frac{1}{3}$

Using the quadratic equation formula,
$x^2-\left(Sumofroot\right)x+\left(Productofroot\right)=0$
Substitute the value in the formula, we get
$x^2-\sqrt{2}x+\frac{1}{3}=0$
Multiply by 3 to remove denominator,
$3x^2-3\sqrt{2}x+1=0$

(iii) 0, $\sqrt{5}$

Using the quadratic equation formula,
$x^2-\left(Sumofroot\right)x+\left(Productofroot\right)=0$
Substitute the value in the formula, we get
$x^2-0x+\sqrt{5}=0$
$x^2+\sqrt{5}=0$

(iv) 1, 1

Using the quadratic equation formula,
$x^2-\left(Sumofroot\right)x+\left(Productofroot\right)=0$
Substitute the value in the formula, we get
$x^2-1x+1=0$
$x^2-x+1=0$

(v) $\frac{-1}{4},\frac{1}{4}$

Using the quadratic equation formula,
$x^2-\left(Sumofroot\right)x+\left(Productofroot\right)=0$
Substitute the value in the formula, we get
$x^2-\frac{-1}{4}x+\frac{1}{4}=0$
Multiply by $4$
$4x^2+x+1=0$

(vi) 4, 1

Using the quadratic equation formula,
$x^2-\left(Sumofroot\right)x+\left(Productofroot\right)=0$
Substitute the value in the formula, we get
$x^2-4x+1=0$