Question

Find the quadratic polynomial, each with the given numbers as the sum and product of its zeroes, respectively.
a) $\dfrac{1}{4},-1$
b) $0,\sqrt{5}$
c) 1,1

Answer 1

Hint: Use the relation between the zeroes and the coefficients of a polynomial. Also, try to draw some results seeing the nature of the roots.

Complete step-by-step answer:

We know, for standard quadratic equation $a{{x}^{2}}+bx+c=0$ , the roots are given by:
$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$

Also, the relation between the coefficients and the roots of a general quadratic equation comes out to be:

Sum of the roots of quadratic equation = $\dfrac{-\left( \text{coefficient of x} \right)}{\text{coefficient of }{{\text{x}}^{2}}}=\dfrac{-b}{a}$ .

Product of the roots of quadratic equation = \[\dfrac{\text{constant term}}{\text{coefficient of }{{\text{x}}^{2}}}\text{=}\dfrac{\text{c}}{\text{a}}\] .

Now a quadratic equation with its roots given can be written as:

$k\left( {{x}^{2}}-\left( \text{sum of the roots} \right)x+\left( \text{product of roots} \right) \right)=0$

Where k is any constant and can have any value according to our need.

a) The quadratic equation with sum and product of its zeroes as $\dfrac{1}{4},-1$ , respectively is:

 $k\left( {{x}^{2}}-\dfrac{1}{4}x-1 \right)=0$

Now, as no condition given in the question, we let k to be 4. Therefore, the quadratic equation with the sum and product of its zeroes as $\dfrac{1}{4},-1$ , respectively is $4{{x}^{2}}-x-4=0$ .

b) The quadratic equation with sum and product of its zeroes as $0,\sqrt{5}$ , respectively is:
 $k\left( {{x}^{2}}-0x+\sqrt{5} \right)=0$

Now, as no condition given in the question, we let k to be 1. Therefore, the quadratic equation with the sum and product of its zeroes as $0,\sqrt{5}$ , respectively is ${{x}^{2}}+\sqrt{5}=0$ .

c) The quadratic equation with sum and product of its zeroes as $1,1$ , respectively is:
 $k\left( {{x}^{2}}-1x+1 \right)=0$

Now, as no condition given in the question, we let k to be 1. Therefore, the quadratic equation with the sum and product of its zeroes as $1,1$, respectively is ${{x}^{2}}-x+1=0$.

Note: Remember that we can form infinite quadratic equations with both of its roots given, just we need to do is to keep on changing the value of k in the equation$k\left( {{x}^{2}}-\left( \text{sum of the roots} \right)x+\left( \text{product of roots} \right) \right)=0$, provided no constraint about the quadratic equation is mentioned in the question.