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Question

Answers

A.a = 1, b = 3

B.a = 3, b = 1

C.a = -1, b = 5

D.a = 5, b = -1

Answer
Verified

Hint: In this problem, we are required to find a and b. So, we need two equations to solve for a and b. One of the equations is given as a + b = 4. Another equation is formed by using the property of factor as factor divides the number completely. By putting x = -2, we get the second equation and our final values are obtained by substitution.

Complete step-by-step answer:

In algebra, a quadratic function is a polynomial function with one or more variables in which the highest-degree term is of the second degree. A single-variable quadratic function can be stated as:

$f(x)=a{{x}^{2}}+bx+c,\quad a\ne 0$

A factor is an expression which divides the whole expression completely without leaving remainder. According to our problem, we are given that x + 2 is a factor of ${{x}^{2}}+ax+2b=0$. So, replacing x = -2, we get

$\begin{align}

& f(x)={{x}^{2}}+ax+2b \\

& f\left( -2 \right)={{\left( -2 \right)}^{2}}+a\left( -2 \right)+2b \\

& f\left( -2 \right)=4-2a+2b \\

& 4-2a+2b=0 \\

& 2b=2a-4 \\

& b=a-2\ldots (1) \\

\end{align}$

And equation (2), is given as a + b =4. Now, putting the value of b in equation (2), we get

$\begin{align}

& a+a-2=4 \\

& 2a=4+2 \\

& a=\dfrac{6}{2} \\

& a=3 \\

& b=3-2 \\

& b=1 \\

\end{align}$

So, the obtained values are a = 3 and b = 1.

Therefore, option (b) is correct.

Note:The key concept involved in solving this problem is the knowledge of factors of an expression. Students must be careful while replacing the value of x. They should replace x = - 2 as x + 2 = 0 and not x = 2.

Complete step-by-step answer:

In algebra, a quadratic function is a polynomial function with one or more variables in which the highest-degree term is of the second degree. A single-variable quadratic function can be stated as:

$f(x)=a{{x}^{2}}+bx+c,\quad a\ne 0$

A factor is an expression which divides the whole expression completely without leaving remainder. According to our problem, we are given that x + 2 is a factor of ${{x}^{2}}+ax+2b=0$. So, replacing x = -2, we get

$\begin{align}

& f(x)={{x}^{2}}+ax+2b \\

& f\left( -2 \right)={{\left( -2 \right)}^{2}}+a\left( -2 \right)+2b \\

& f\left( -2 \right)=4-2a+2b \\

& 4-2a+2b=0 \\

& 2b=2a-4 \\

& b=a-2\ldots (1) \\

\end{align}$

And equation (2), is given as a + b =4. Now, putting the value of b in equation (2), we get

$\begin{align}

& a+a-2=4 \\

& 2a=4+2 \\

& a=\dfrac{6}{2} \\

& a=3 \\

& b=3-2 \\

& b=1 \\

\end{align}$

So, the obtained values are a = 3 and b = 1.

Therefore, option (b) is correct.

Note:The key concept involved in solving this problem is the knowledge of factors of an expression. Students must be careful while replacing the value of x. They should replace x = - 2 as x + 2 = 0 and not x = 2.

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