Question

find the sum and product of roots of the quadratic equation 3x^2 + 7x + 1 = 0​

Answer 1

Step-by-step explanation:

Comparing 3x² + 7x + 1 = 0 and ax² + bx + c = 0

we get

a = 3, b = 7 and c = 1

Therefore,

Sum of the roots = -b/a = -7/3

Product of the roots = c/a = 1/3

Answer 2

[ Method 1 ]

Here the given quadratic equation is

$\quad 3x^{2}+7x+1=0$

Comparing with the general quadratic equation $ax^{2}+bx+c=0$, we get

$\quad a=3,b=7,c=1$

Now using quadratic formula, we get

$\quad x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$

$\Rightarrow x=\dfrac{-7\pm\sqrt{49-12}}{6}$

$\Rightarrow x=\dfrac{-7\pm\sqrt{37}}{6}$

So the roots of the given equation are

$\quad x=\dfrac{-7+\sqrt{37}}{6},\:x=\dfrac{-7-\sqrt{37}}{6}$

Required sum

$=\dfrac{-7+\sqrt{37}}{6}+\dfrac{-7-\sqrt{37}}{6}$

$=\dfrac{-7+\sqrt{37}-7-\sqrt{37}}{6}$

$=\dfrac{-14}{6}$

$=\dfrac{-7}{3}$

Required product

$=\dfrac{-7+\sqrt{37}}{6}\times\dfrac{-7-\sqrt{37}}{6}$

$=\dfrac{49-37}{36}$

$=\dfrac{12}{36}$

$=\dfrac{1}{3}$

[ Method 2 ]

We know that, if $ax^{2}+bx+c=0$ where $a\neq 0$ be a quadratic equation, then

• sum of the roots $=\dfrac{-b}{a}$
• product of the roots $=\dfrac{c}{a}$

Comparison has already given us: $a = 3, b = 7, c = 1$

Thus, sum of the roots $=\dfrac{-7}{3}$

and product of the roots $=\dfrac{1}{3}$

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