Question

if one root of the quadratic polynomial 2 x square - 3 x + P is 3 find the other route also find the value of p

Answer 1

$2 {x}^{2} - 3x + p \\ 2( {3})^{2} - 3(3) + p = 0 \\ 18 - 9 + p = 0 \\ p = - 9 \\ 2 {x}^{2} - 3x - 9 = 0 \\ 2 {x}^{2} - 6x + 3x - 9 = 0 \\ 2x(x - 3) + 3(x - 3) = 0 \\ (x - 3)(2x + 3) = 0 \\ 2x + 3 = 0 \\ x = \frac{ - 3}{2}$
another root is -3/2, p= -9

Answer 2

Answer:

Other root is -1/2 and the value of p is -9.

Step-by-step explanation:

Let,

$f(x) =2x^2-3x+p$

Since, 3 is a root of f(x),

That is, f(3) = 0,

$2(3)^2-3(3)+p=0$

$18-9+p=0$

$9+p=0$

$\implies p=-9$

Thus, the required quadratic equation is,

$f(x) =2x^2-3x-9$

At the roots of f(x),

$f(x)=0$

$2x^2-3x-9=0$

$2x^2-6x+3x-9=0$

$2x(x-3)+1(3x-9)=0$

$(2x+1)(3x-9)=0$

⇒ 2x+1 = 0 or 3x-9=0

⇒ x = -1/2 or x = 3,

Hence, if 3 is one root of f(x) then other root is -1/2.