Question

If both x+2 and 2x+1are factors of ax^2+2x+b then the value of a-b is

Answer 1

Answer:

a=4\5 and b=4/5

Step-by-step explanation:

we can get it by putting two values of x and solve the two linear equations we get

Answer 2

The value of a-b is 0

That is a-b=0

The values of a and b is $\frac{4}{5}$ and $\frac{4}{5}$

Step-by-step explanation:

Given that  x+2 and 2x+1 are the factors of the quadratic polynomial expression $ax^2+2x+b$

Let P(x)=$ax^2+2x+b$

Since x+2 and 2x+1 are factors then the zeros are x+2=0 and 2x+1=0

x=-2 and $x=-\frac{1}{2}$

To find the value of a-b :

• First find the values of a and b

Since x=-2 is the zero of P(x)

Therefore P(-2)=0

Put x=-2 in the given expression $P(x)=ax^2+2x+b$

$P(-2)=a(-2)^2+2(-2)+b=0$

$4a-4+b=0$

$4a+b=4\hfill (1)$

Since $x=-\frac{1}{2}$ is the zero of P(x)

Therefore P($-\frac{1}{2}$)=0

Put $x=-\frac{1}{2}$ in the given expression $P(x)=ax^2+2x+b$

$P(-\frac{1}{2})=a(-\frac{1}{2})^2+2(-\frac{1}{2})+b=0$

$a(\frac{1}{4})-1+b=0$

$\frac{a-4+4b}{4}=0$

$a-4+4b=0$

$a+4b=4\hfill (2)$

• Now solving the equations (1) and (2)

Multiply the equation (1) into 4 we get

$16a+4b=16\hfill (3)$

Now subtracting the equations (2) and (3)

$a+4b=4$

$16a+4b=16$

___(-)__(-)__(-)_________

$-15a=-12$

$=\frac{12}{15}$

Therefore $a=\frac{4}{5}$

• Now substitute the value of $a=\frac{4}{5}$ in equation (1)

$4(\frac{4}{5})+b=4$

$\frac{16}{5}+b=4$

$b=4-\frac{16}{5}$

$=\frac{20-16}{5}$

Therefore $b=\frac{4}{5}$

Therefore the values of a and b is $\frac{4}{5}$ and $\frac{4}{5}$

Now to find a-b

Substitute the values of a and b in the above expression we get

$\frac{4}{5}-\frac{4}{5}=0$

Therefore a-b=0