If one root of the quadratic equation 2x^{2} | vedclass

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(A) The given quadratic equation is $2x^{2} + Px + 4 = 0$.Since $2$ is a root of the equation,it must satisfy the equation.Substituting $x = 2$ in the

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$1, -6$

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(A) The given quadratic equation is $2x^{2} + Px + 4 = 0$.Since $2$ is a root of the equation,it must satisfy the equation.Substituting $x = 2$ in the equation:$2(2)^{2} + P(2) + 4 = 0$$2(4) + 2P + 4 = 0$$8 + 2P + 4 = 0$$2P + 12 = 0$$2P = -12$$P = -6$.Now,substitute $P = -6$ back into the original equation:$2x^{2} - 6x + 4 = 0$.Divide the entire equation by $2$:$x^{2} - 3x + 2 = 0$.Factorizing the quadratic equation:$x^{2} - 2x - x + 2 = 0$$x(x - 2) - 1(x - 2) = 0$$(x - 1)(x - 2) = 0$.So,the roots are $x = 1$ and $x = 2$.Since one root is given as $2$,the second root is $1$.Thus,the second root is $1$ and the value of $P$ is $-6$.

If one root of the quadratic equation $2x^{2} + Px + 4 = 0$ is $2$,find the second root and the value of $P$.

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