Question

which of the following has -1/2 and 2/5 as their zeroes
A) 5x2-4x+5
B) 2x2-x+10
C) 10x2-7x+1
D) 10x2+x-2

answer with solutions
​

Answer 1

We know the equation to find the zeros,

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

Lets look into the options:

__________

(A) 5x² - 4x + 5 =0

[ a=5, b=-4, c=5 ]

Applying the values to the equation:

$\sf{ \implies x=\dfrac{-(-4)\pm \sqrt{(-4)^2-(4\times 5\times 5)}}{2\times 5}}$

$\sf{ \implies x=\dfrac{4\pm \sqrt{16-100}}{10}}$

$\sf{ \implies x=\dfrac{4\pm \sqrt{-84}}{10}}$

It is clear that 5x² - 4x + 5 =0 don't have real roots since negative of a root is not defined.

__________

(B) 2x² - x + 10 =0

[ a=2, b=-1, c=10 ]

Applying the values to the equation:

$\sf{ \implies x=\dfrac{-(-1)\pm \sqrt{(-1)^2-(4\times 2\times 10)}}{2\times 2}}$

$\sf{ \implies x=\dfrac{1\pm \sqrt{1-80}}{4}}$

$\sf{ \implies x=\dfrac{1\pm \sqrt{-79}}{4}}$

It is clear that 2x² - x + 10 =0 don't have real roots since negative of a root is not defined.

__________

(C) 10x² - 7x + 1 =0

[ a=10, b=-7, c=1 ]

Applying the values to the equation:

$\sf{ \implies x=\dfrac{-(-7)\pm \sqrt{(-7)^2-(4\times 10\times 1)}}{2\times 10}}$

$\sf{ \implies x=\dfrac{7\pm \sqrt{49-40}}{20}}$

$\sf{ \implies x=\dfrac{7\pm \sqrt{9}}{20}}$

$\sf{ \implies x=\dfrac{7\pm 3}{20}}$

$\sf{ \implies x=\dfrac{7+ 3}{20} \ \ \ \& \ \ \ x=\dfrac{7- 3}{20}}$

$\sf{ \implies x=\dfrac{10}{20} \ \ \ \& \ \ \ x=\dfrac{4}{20}}$

$\sf{ \implies x=\dfrac{1}{2} \ \ \ \& \ \ \ x=\dfrac{1}{5}}$

∴ zeroes of 10x²-7x+1 are  $\sf{\frac{1}{2} \ and \ \frac{1}{5}}$.

__________

(D) 10x² + x - 2 =0

[ a=10, b=1, c=-2 ]

Applying the values to the equation:

$\sf{ \implies x=\dfrac{(-1)\pm \sqrt{(1)^2-(4\times 10\times -2)}}{2\times 10}}$

$\sf{ \implies x=\dfrac{-1\pm \sqrt{1+80}}{20}}$

$\sf{ \implies x=\dfrac{-1\pm \sqrt{81}}{20}}$

$\sf{ \implies x=\dfrac{-1\pm 9}{20}}$

$\sf{ \implies x=\dfrac{-1+ 9}{20} \ \ \ \& \ \ \ x=\dfrac{-1- 9}{20}}$

$\sf{ \implies x=\dfrac{8}{20} \ \ \ \& \ \ \ x=\dfrac{-10}{20}}$

$\sf{ \implies x=\dfrac{2}{5} \ \ \ \& \ \ \ x=\dfrac{-1}{2}}$

∴ zeroes of 10x²-7x+1 are  $\boxed{\sf{\frac{-1}{2} \ and \ \frac{2}{5}}}$.  [Which is he answer required]

__________

∴ Answer = (D) 10x² + x - 2

Answer 2

Answer:

option D is the answer

Step-by-step explanation:

option D is the answer