Question

8
If one zero of the quadratic
polynomial 2x^2-3x+p is 3,find the
other zero,find the value of p
A.3/2,-9
B.-3/2,-9
C.3/2,9
D.-3/2,9​

Answer 1

Solution :

$\bf{\red{\underline{\bf{Given\::}}}}$

If one zero of the quadratic polynomial 2x² - 3x + p is 3.

$\bf{\red{\underline{\bf{To\:find\::}}}}$

The other zero and the value of p.

$\bf{\red{\underline{\bf{Explanation\::}}}}$

We have f(x) = 2x² - 3x + p

Zero of the polynomial f(x) = 0

3 is the zero of f(x)

So;

$\longrightarrow\sf{2(3)^{2} -3(3)+p=0}\\\\\longrightarrow\sf{2\times 9-9+p=0}\\\\\longrightarrow\sf{18-9+p=0}\\\\\longrightarrow\sf{9+p=0}\\\\\longrightarrow\sf{p=0-9}\\\\\longrightarrow\sf{\orange{p=-9}}$

Now;

$\longrightarrow\sf{f(x)=2x^{2} -3x-9=0}\\\\\longrightarrow\sf{2x^{2} -6x+3x-9=0}\\\\\longrightarrow\sf{2x(x-3)+3(x-3)=0}\\\\\longrightarrow\sf{(x-3)(2x+3)=0}\\\\\longrightarrow\sf{x-3=0\:\:\:Or\:\:\:2x+3=0}\\\\\longrightarrow\sf{x=3\:\:\:Or\:\:\:2x=-3}\\\\\longrightarrow\sf{\orange{x=3\:\:\:Or\:\:\:x=\dfrac{-3}{2} }}$

Thus;

$\underbrace{\bf{The\:other\:zero\:will\:be\:\:\boxed{\sf{\frac{-3}{2} }}}}}\\\underbrace{\bf{The\:value \:of\:p\:wil\:be\:\:\boxed{\sf{-9 }}}}}$

Answer 2

Answer: Opt.B

[One zero of the quadratic polynomial is -3/2;

Value of p is -9.]

Given polynomial: 2x² - 3x + p = 0

.°. f(x) = 2x² - 3x + p

Given that, 3 is the zero of f(x),

So, f(3) = 0

Putting the value as 3, in place of x,

=> 2(3)² - 3(3) + p = 0

=> 2 × 9 - 9 + p = 0

=> 18 - 9 + p = 0

=> p = -9

Thus, we got the value of p as -9.

Putting it, in the given polynomial:

f(x) = 2x² - 3x + (-9)

Splitting the middle term,

= 2x² - 6x + 3x - 9

= 2x(x - 3) + 3(x - 3)

= (2x + 3) (x - 3)

Now, When f(x) = 0

(2x + 3) (x - 3) = 0

2x + 3 = 0 or x - 3 = 0

x = -3/2 or x = 3

As Opt. B matches our solution,

So, x = -3/2.

Thus, the other zero of the given polynomial is -3/2 and value of p is -9. [Opt. B: -3/2, -9]