
Find the quadratic equation whose zeroes are –3 and 4?
(a) $ {{x}^{2}}-x+12 $
(b) $ {{x}^{2}}+x+12 $
(c) $ \dfrac{{{x}^{2}}}{2}-\dfrac{x}{2}-6 $
(d) $ 2{{x}^{2}}+2x-24 $
Answer
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Hint: We start solving the problem by recalling the fact that the quadratic polynomial with roots a and b is defined as $ \left( x-a \right)\left( x-b \right) $ . We use this fact for the required polynomial and perform the necessary calculations. We then recall the fact that dividing or multiplying a polynomial will not alter its zeros. We then divide the obtained quadratic polynomial with 2 to get the required answer.
Complete step by step answer:
According to the problem, we are asked to find the quadratic polynomial whose roots are given as –3 and 4.
We know that the quadratic polynomial with roots a and b is defined as $ \left( x-a \right)\left( x-b \right) $ .
So, the required quadratic polynomial is $ \left( x-\left( -3 \right) \right)\left( x-4 \right) $ .
$ \Rightarrow \left( x-\left( -3 \right) \right)\left( x-4 \right)=\left( x+3 \right)\left( x-4 \right) $ .
$ \Rightarrow \left( x-\left( -3 \right) \right)\left( x-4 \right)={{x}^{2}}-4x+3x-12 $ .
$ \Rightarrow \left( x-\left( -3 \right) \right)\left( x-4 \right)={{x}^{2}}-x-12 $ ---(1).
We know that dividing or multiplying a polynomial will not alter its zeros.
So, let us divide the quadratic polynomial in equation (1) with 2.
So, we get quadratic polynomial as $ \dfrac{{{x}^{2}}-x-12}{2}=\dfrac{{{x}^{2}}}{2}-\dfrac{x}{2}-6 $ .
We have found the required quadratic polynomial is $ \dfrac{{{x}^{2}}}{2}-\dfrac{x}{2}-6 $ .
$ \therefore $ The correct option for the given problem is (c).
Note:
We can also solve this problem by using the fact that the quadratic polynomial with zeros a and b as $ {{x}^{2}}-\left( a+b \right)x+\left( a\times b \right) $ . We should perform each step carefully in order to avoid calculation mistakes and confusion. We should not stop solving the problem after finding the quadratic polynomial as $ {{x}^{2}}-x-12 $ . Similarly, we can expect problems to find the quadratic polynomial if the given zeros are increased by 1.
Complete step by step answer:
According to the problem, we are asked to find the quadratic polynomial whose roots are given as –3 and 4.
We know that the quadratic polynomial with roots a and b is defined as $ \left( x-a \right)\left( x-b \right) $ .
So, the required quadratic polynomial is $ \left( x-\left( -3 \right) \right)\left( x-4 \right) $ .
$ \Rightarrow \left( x-\left( -3 \right) \right)\left( x-4 \right)=\left( x+3 \right)\left( x-4 \right) $ .
$ \Rightarrow \left( x-\left( -3 \right) \right)\left( x-4 \right)={{x}^{2}}-4x+3x-12 $ .
$ \Rightarrow \left( x-\left( -3 \right) \right)\left( x-4 \right)={{x}^{2}}-x-12 $ ---(1).
We know that dividing or multiplying a polynomial will not alter its zeros.
So, let us divide the quadratic polynomial in equation (1) with 2.
So, we get quadratic polynomial as $ \dfrac{{{x}^{2}}-x-12}{2}=\dfrac{{{x}^{2}}}{2}-\dfrac{x}{2}-6 $ .
We have found the required quadratic polynomial is $ \dfrac{{{x}^{2}}}{2}-\dfrac{x}{2}-6 $ .
$ \therefore $ The correct option for the given problem is (c).
Note:
We can also solve this problem by using the fact that the quadratic polynomial with zeros a and b as $ {{x}^{2}}-\left( a+b \right)x+\left( a\times b \right) $ . We should perform each step carefully in order to avoid calculation mistakes and confusion. We should not stop solving the problem after finding the quadratic polynomial as $ {{x}^{2}}-x-12 $ . Similarly, we can expect problems to find the quadratic polynomial if the given zeros are increased by 1.
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