If one root of a quadratic equation 3x2- PX - 4 - 0 is 2-3- find the value of p

The given equation is nbsp;3x2 + px + 4 = 0since x = 2/3 is a root of the above equation, then it must satisfy it. nbsp;3(2/3)2 + 2p/3 + 4 = 0â‡’ 4/3 + 4 + ...

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Standard XII

Mathematics

Roots of a Quadratic Equation

If one root o...

Question

If one root of a quadratic equation $3 x^{2} + P x + 4 = 0$ is $\frac{2}{3}$ , find the value of $P$ .

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Solution

We have,
$3 x^{2} + P x + 4 = 0$
$\frac{2}{3}$ is the root of the above equation.
So,
$3 (\frac{2}{3})^{2} + \frac{2}{3} P + 4 = 0$
$\Rightarrow$ $\frac{4}{3} + \frac{2}{3} P + 4 = 0$
$\Rightarrow$ $\frac{2}{3} P = \frac{- 16}{3}$
$\Rightarrow$ $P = - 8$
Hence, the value of $P$ is $- 8$ .

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If one root of a quadratic equation 3x2+Px+4=0 is 23, find the value of P.

We have,3x2+Px+4=0 23 is the root of the above equation.So,3(23)2+23P+4=0⇒ 43+23P+4=0⇒ 23P=−163⇒ P=−8Hence, the value of P is −8.

If one root of a quadratic equation $3 x^{2} + P x + 4 = 0$ is $\frac{2}{3}$ , find the value of $P$ .

We have,
$3 x^{2} + P x + 4 = 0$
$\frac{2}{3}$ is the root of the above equation.
So,
$3 (\frac{2}{3})^{2} + \frac{2}{3} P + 4 = 0$
$\Rightarrow$ $\frac{4}{3} + \frac{2}{3} P + 4 = 0$
$\Rightarrow$ $\frac{2}{3} P = \frac{- 16}{3}$
$\Rightarrow$ $P = - 8$
Hence, the value of $P$ is $- 8$ .