
If one of the zeros of the quadratic polynomial $ 2{{x}^{2}}+px+4 $ is 2, find the other zero of the given quadratic polynomial. Also find the value of p.
A.p = -1, other zero = 1
B.p = -2, other zero = 1
C.p = -6, other zero = 1
D.p = -7, other zero = 1
Answer
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Hint: Here, we know the formula that if $ m $ and $ n $ are two roots of the quadratic equation of the form $ a{{x}^{2}}+bx+c=0 $ then the sum and product of the roots of $ m $ and $ n $ can be written as:
$ m+n=\dfrac{-b}{a} $ and $ m.n=\dfrac{c}{a} $ . We shall use these two relations to obtain the answer.
Complete step-by-step answer:
In the question, the quadratic polynomial is given as $ 2{{x}^{2}}+px+4 $ . So the corresponding quadratic equation is:
$ 2{{x}^{2}}+px+4=0\text{ }.......\text{ (1)} $
We know that for a quadratic equation there will be two roots. Here one root is given as 2 and we need to find the other.
We know that for a quadratic equation of the form $ a{{x}^{2}}+bx+c=0 $ , the product of the roots is $ \dfrac{c}{a}. $ i.e. let m and n be two roots of the equation then:
$ mn=\dfrac{c}{a} $
Now, consider equation (1) where a = 2, b = p and c = 4 and also given that one root, m=2. Next we have to find n.
For that we have,
$ \begin{align}
& 2n=\dfrac{4}{2} \\
& 2n=2 \\
& n=1 \\
\end{align} $
We got the other root n = 1.
Therefore, the two roots of the quadratic equation (1) are 2, 1.
So, next we have to find the value of p.
We know that for a quadratic equation of the form $ \text{ }a{{x}^{2}}+bx+c=0 $ , the sum of the roots is $ \dfrac{-b}{a}. $
Hence, for the two roots m and n we get:
$ \begin{align}
& m+n=\dfrac{-b}{a} \\
& 2+1=\dfrac{-p}{2} \\
& 3=\dfrac{-p}{2} \\
& \\
\end{align} $
By cross multiplication we get, $ $
$ 6=-p $
Therefore, we can say that the value of p = -6.
We got the value of the other root as 1 and p = -6.
Hence, the correct answer for this question is option (c).
Note: Here an alternate method to find the value of p is by using the factors (x-1) and (x-2). That is by multiplying them and equating them to zero we will get the quadratic equation. From the equation you can find the value of p.
$ m+n=\dfrac{-b}{a} $ and $ m.n=\dfrac{c}{a} $ . We shall use these two relations to obtain the answer.
Complete step-by-step answer:
In the question, the quadratic polynomial is given as $ 2{{x}^{2}}+px+4 $ . So the corresponding quadratic equation is:
$ 2{{x}^{2}}+px+4=0\text{ }.......\text{ (1)} $
We know that for a quadratic equation there will be two roots. Here one root is given as 2 and we need to find the other.
We know that for a quadratic equation of the form $ a{{x}^{2}}+bx+c=0 $ , the product of the roots is $ \dfrac{c}{a}. $ i.e. let m and n be two roots of the equation then:
$ mn=\dfrac{c}{a} $
Now, consider equation (1) where a = 2, b = p and c = 4 and also given that one root, m=2. Next we have to find n.
For that we have,
$ \begin{align}
& 2n=\dfrac{4}{2} \\
& 2n=2 \\
& n=1 \\
\end{align} $
We got the other root n = 1.
Therefore, the two roots of the quadratic equation (1) are 2, 1.
So, next we have to find the value of p.
We know that for a quadratic equation of the form $ \text{ }a{{x}^{2}}+bx+c=0 $ , the sum of the roots is $ \dfrac{-b}{a}. $
Hence, for the two roots m and n we get:
$ \begin{align}
& m+n=\dfrac{-b}{a} \\
& 2+1=\dfrac{-p}{2} \\
& 3=\dfrac{-p}{2} \\
& \\
\end{align} $
By cross multiplication we get, $ $
$ 6=-p $
Therefore, we can say that the value of p = -6.
We got the value of the other root as 1 and p = -6.
Hence, the correct answer for this question is option (c).
Note: Here an alternate method to find the value of p is by using the factors (x-1) and (x-2). That is by multiplying them and equating them to zero we will get the quadratic equation. From the equation you can find the value of p.
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