Question

Find the zeros of the following quadratic polynomials and verify the relationship between the zeros and the coefficients.
(i) $6{{x}^{2}}-3-7x$
(ii) $4{{u}^{2}}+8u$

Answer 1

Hint: To find the zeros of the given quadratic polynomials and verify the relationship between the zeros and the coefficients, we will write the polynomial in the form $a{{x}^{2}}+bx+c=0$ . Find the zeros by using either the splitting of the middle terms or $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ or solving by taken the common factors out. After finding the zeros verify the relationship between the zeros and the coefficients using the formulas $\text{Sum of zeros }=-\dfrac{\text{Coefficient of }x}{\text{Coefficient of }{{x}^{2}}}$ and $\text{Product of zeros}=\dfrac{\text{Constant term}}{\text{Coefficient of }{{x}^{2}}}$ .

Complete step by step answer:
We have to find the zeros of the given quadratic polynomials and verify the relationship between the zeros and the coefficients.
(i) Let us find the zeros of $6{{x}^{2}}-3-7x$ .
Let us write this in the form $a{{x}^{2}}+bx+c=0$ . That is,
$6{{x}^{2}}-7x-3=0$
Let us solve this by using the formula $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
From the given quadratic polynomial, we can write
$a=6,b=-7,c-3$
Hence,
$x=\dfrac{-\left( -7 \right)\pm \sqrt{{{\left( -7 \right)}^{2}}-\left( 4\times 6\times -3 \right)}}{2\times 6}$
This can be written as
$x=\dfrac{7\pm \sqrt{49-4\times 6\times -3}}{2\times 6}$
Let us solve this. We will get
$x=\dfrac{7\pm \sqrt{49+72}}{12}$
By adding the terms inside the roots, we will get
$x=\dfrac{7\pm \sqrt{121}}{12}$
Let us take the square root of 121. We will get
$x=\dfrac{7\pm 11}{12}$
We can write the above form as
$\begin{align}
  & x=\dfrac{7+11}{12},x=\dfrac{7-11}{12} \\
 & \Rightarrow x=\dfrac{18}{12},x=\dfrac{-4}{12} \\
\end{align}$
When we simplify this, we will get
$x=\dfrac{3}{2},\dfrac{-1}{3}$
Hence, the roots of $6{{x}^{2}}-3-7x$ are $\dfrac{3}{2},\dfrac{-1}{3}$ .
Now, we have to verify the relationship between the roots and the coefficients.
We know that $\text{Sum of zeros }=-\dfrac{\text{Coefficient of }x}{\text{Coefficient of }{{x}^{2}}}$ and $\text{Product of zeros}=\dfrac{\text{Constant term}}{\text{Coefficient of }{{x}^{2}}}$ .
Let us first consider $\text{Sum of zeros }=-\dfrac{\text{Coefficient of }x}{\text{Coefficient of }{{x}^{2}}}$ .
Consider the LHS.
Sum of zeros $=\dfrac{3}{2}+\dfrac{-1}{3}$
Taking the LCM =6 and solving, we will get
Sum of zeros $=\dfrac{9-2}{6}=\dfrac{7}{6}$
Now let us see the RHS.
$-\dfrac{\text{Coefficient of }x}{\text{Coefficient of }{{x}^{2}}}=-\dfrac{-7}{6}=\dfrac{7}{6}$
We can see that LHS=RHS. Hence, verified.
Now, let us consider $\text{Product of zeros}=\dfrac{\text{Constant term}}{\text{Coefficient of }{{x}^{2}}}$ .
We can first find the LHS. We will get
$\text{Product of zeros}=\dfrac{3}{2}\times \dfrac{-1}{3}=-\dfrac{1}{2}$
Let us consider the RHS. We get
$\dfrac{\text{Constant term}}{\text{Coefficient of }{{x}^{2}}}=\dfrac{-3}{6}=\dfrac{-1}{2}$
We can see that LHS=RHS. Hence, we verified the relationship between the zeros and coefficients.
(ii) Let us find the roots of $4{{u}^{2}}+8u$ . Let us equate it to 0. That is
$4{{u}^{2}}+8u=0$
We can now take the common terms outside. We will get
$4u\left( u+2 \right)=0$
This means that either $4u=0\text{ or }u+2=0$
Let us consider $4u=0$
$\Rightarrow u=0$
Now, let us consider $u+2=0$
$\Rightarrow u=-2$
Hence, the zeros are 0 and -2.
Now, let us verify the relationship between the roots and the coefficients.
We know that $\text{Sum of zeros }=-\dfrac{\text{Coefficient of }x}{\text{Coefficient of }{{x}^{2}}}$ and $\text{Product of zeros}=\dfrac{\text{Constant term}}{\text{Coefficient of }{{x}^{2}}}$ .
Let us first consider $\text{Sum of zeros }=-\dfrac{\text{Coefficient of }x}{\text{Coefficient of }{{x}^{2}}}$ .
We will consider the LHS.
$\text{Sum of zeros }=0-2=-2$
Now, we can consider the RHS.
$-\dfrac{\text{Coefficient of }x}{\text{Coefficient of }{{x}^{2}}}=-\dfrac{8}{4}=-2$
We can see that LHS=RHS. Hence, verified.
Now, let us consider $\text{Product of zeros}=\dfrac{\text{Constant term}}{\text{Coefficient of }{{x}^{2}}}$ .
We can first find the LHS. We will get
$\text{Product of zeros}=0\times -2=0$
Let us consider the RHS. We get
$\dfrac{\text{Constant term}}{\text{Coefficient of }{{x}^{2}}}=\dfrac{0}{4}=0$
We can see that LHS=RHS. Hence, we verified the relationship between the zeros and coefficients.

Note: You can use any method to find the roots of a quadratic equation. There can be a chance of mistake when writing the formula for verification as $\text{Sum of zeros }=-\dfrac{\text{Coefficient of }{{x}^{2}}}{\text{Coefficient of }x}$ , \[\text{Sum of zeros }=\dfrac{\text{Constant term}}{\text{Coefficient of }{{x}^{2}}}\] , $\text{Product of zeros}=\dfrac{\text{Coefficient of }{{x}^{2}}}{\text{Constant term}}$ or $\text{Product of zeros}=-\dfrac{\text{Coefficient of }x}{\text{Coefficient of }{{x}^{2}}}$ . The roots found must satisfy the verification formulas $\text{Sum of zeros }=-\dfrac{\text{Coefficient of }x}{\text{Coefficient of }{{x}^{2}}}$ and $\text{Product of zeros}=\dfrac{\text{Constant term}}{\text{Coefficient of }{{x}^{2}}}$ , else the found roots are wrong.