Question

3 and 1 /3 are the roots of quadratic equation then the quadratic equation is??

Answer 1

If $3$ and $\frac{1}{3}$ are the roots of a quadratic equation then the quadratic equation is $3x^{2} +10x+1$.

Given:

$3$ and $\frac{1}{3}$ are the roots of a quadratic equation.

To find:

The quadratic equation whose roots are given.

Step by step explanation:

Let $\alpha=3,\beta=\frac{1}{3}$,

Then the required quadratic equation is,$x^{2} -(\alpha+\beta)x+\alpha\beta$.

$\alpha +\beta=3+\frac{1}{3}=\frac{10}{3}$

$\alpha\cdot\beta=3\cdot\frac{1}{3} =1$

Therefore,

$x^{2} -\frac{10}{3}x+1$

multiplying by $3$, we get

$3x^{2} +10x+1$

Hence, $3x^{2} +10x+1$ is the required quadratic equation whose roots are $3$  and $\frac{1}{3}$.

Answer 2

Step-by-step explanation:

As per data given in the question,

We have,

$3 \:and \:\frac{1}{3}$ are the roots of the quadratic equation.

Let ∝ and β are the roots of the equation.

Hence, the quadratic equation will be

$(x - \alpha) (x - \beta)$

So, we can write it as,

$(x-3) (x-\frac{1}{3})=0\\\Rightarrow (x-3) (\frac{3x-1}{3})=0\\\Rightarrow (x-3) (3x-1)=0\\\Rightarrow (x \times 3x) -(x \times 1) - (3 \times 3x) + (-3 \times -1)=0\\\Rightarrow3x^2-x-9x+3=0\\\Rightarrow 3x^2-10x+3=0$

Answer: Hence, the required quadratic equation will be $3x^2-10x+3=0$