Answer
Verified
477.6k+ views
Hint: First of all, compare the given quadratic equation by general quadratic equation \[a{{x}^{2}}+bx+c=0\] and get the values of a, b and c. Then find the zeros of the equation by using
\[X=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\text{ or }X=\dfrac{-b\pm \sqrt{D}}{2a}\]
Complete step-by-step answer:
Here we have to find the zeros of a quadratic polynomial \[\left( 6{{x}^{2}}-7x-3 \right)\]. Before proceeding with this question, we must know what the root of an equation means. First of all, roots or zeros of any equation are the values of the variable which satisfy the equation. In other words, if by substituting some value of the variable, the equation becomes zero, then that value of the variable is the root of the equation. For example, 5x + 10 = 0 has a root – 2 because by substituting x = – 2, this equation becomes 0.
For any general quadratic equation \[a{{x}^{2}}+bx+c=0\] where \[a,b,c\in R\] and \[a\ne 0\], then, the solution or zeroes of the quadratic equation is given by:
\[X=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]
That means, \[X=\dfrac{-b+\sqrt{{{b}^{2}}-4ac}}{2a}\text{ }\]
And \[X=\dfrac{-b-\sqrt{{{b}^{2}}-4ac}}{2a}\text{ }\]
The expression \[{{b}^{2}}-4ac=D\] is called the discriminant of the quadratic equation. So, we can also write zeroes as,
\[X=\dfrac{-b+\sqrt{D}}{2a}\text{ }\]
And, \[X=\dfrac{-b-\sqrt{D}}{2a}\text{ }\]
Now, let us consider the quadratic polynomial given in the question,
\[Q\left( x \right)=6{{x}^{2}}-7x-3=0\]
By comparing the above equation by general quadratic equation \[a{{x}^{2}}+bx+c=0\], we get,
a = 6, b = – 7 and c = – 3
Let us find the discriminant of the above equation, we get,
\[D={{b}^{2}}-4ac\]
By substituting the values of a, b and c, we get,
\[\begin{align}
& D={{\left( -7 \right)}^{2}}-4\left( 6 \right)\left( -3 \right) \\
& =49+72 \\
& D=121 \\
\end{align}\]
We know that the zeroes of the quadratic equation are:
\[X=\dfrac{-b+\sqrt{D}}{2a}\text{ }\]
And, \[X=\dfrac{-b-\sqrt{D}}{2a}\text{ }\]
By substituting the values of a, b and D, we get,
\[X=\dfrac{-\left( -7 \right)+\sqrt{121}}{2\times \left( 6 \right)}\]
And, \[X=\dfrac{-\left( -7 \right)-\sqrt{121}}{2\times \left( 6 \right)}\]
We know that \[\sqrt{121}=11\], so we get,
\[X=\dfrac{7+11}{12}=\dfrac{18}{12}=\dfrac{3}{2}\]
And \[X=\dfrac{7-11}{12}=\dfrac{-4}{12}=\dfrac{-1}{3}\]
So, we get the roots of the equation \[6{{x}^{2}}-7x-3=0\] as \[\dfrac{3}{2}\] and \[\dfrac{-1}{3}\].
Note:
Students can also find the zeroes of the quadratic equation \[6{{x}^{2}}-7x-3\] as follows:
\[6{{x}^{2}}-7x-3=0\]
Here, we can write 7x = 9x – 2x, we get,
\[6{{x}^{2}}-\left( 9x-2x \right)-3=0\]
Or, \[6{{x}^{2}}-9x+2x-3=0\]
We can also write the above equation as,
\[3x\left( 2x-3 \right)+\left( 2x-3 \right)=0\]
By taking (2x – 3) common, we get,
\[\left( 2x-3 \right)\left( 3x+1 \right)=0\]
So, we get \[\left( 2x-3 \right)=0\] and \[\left( 3x+1 \right)=0\]
\[x=\dfrac{3}{2}\text{ and }x=\dfrac{-1}{3}\]
So, we get the roots of the quadratic equation as \[\dfrac{3}{2}\] and \[\dfrac{-1}{3}\].
\[X=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\text{ or }X=\dfrac{-b\pm \sqrt{D}}{2a}\]
Complete step-by-step answer:
Here we have to find the zeros of a quadratic polynomial \[\left( 6{{x}^{2}}-7x-3 \right)\]. Before proceeding with this question, we must know what the root of an equation means. First of all, roots or zeros of any equation are the values of the variable which satisfy the equation. In other words, if by substituting some value of the variable, the equation becomes zero, then that value of the variable is the root of the equation. For example, 5x + 10 = 0 has a root – 2 because by substituting x = – 2, this equation becomes 0.
For any general quadratic equation \[a{{x}^{2}}+bx+c=0\] where \[a,b,c\in R\] and \[a\ne 0\], then, the solution or zeroes of the quadratic equation is given by:
\[X=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]
That means, \[X=\dfrac{-b+\sqrt{{{b}^{2}}-4ac}}{2a}\text{ }\]
And \[X=\dfrac{-b-\sqrt{{{b}^{2}}-4ac}}{2a}\text{ }\]
The expression \[{{b}^{2}}-4ac=D\] is called the discriminant of the quadratic equation. So, we can also write zeroes as,
\[X=\dfrac{-b+\sqrt{D}}{2a}\text{ }\]
And, \[X=\dfrac{-b-\sqrt{D}}{2a}\text{ }\]
Now, let us consider the quadratic polynomial given in the question,
\[Q\left( x \right)=6{{x}^{2}}-7x-3=0\]
By comparing the above equation by general quadratic equation \[a{{x}^{2}}+bx+c=0\], we get,
a = 6, b = – 7 and c = – 3
Let us find the discriminant of the above equation, we get,
\[D={{b}^{2}}-4ac\]
By substituting the values of a, b and c, we get,
\[\begin{align}
& D={{\left( -7 \right)}^{2}}-4\left( 6 \right)\left( -3 \right) \\
& =49+72 \\
& D=121 \\
\end{align}\]
We know that the zeroes of the quadratic equation are:
\[X=\dfrac{-b+\sqrt{D}}{2a}\text{ }\]
And, \[X=\dfrac{-b-\sqrt{D}}{2a}\text{ }\]
By substituting the values of a, b and D, we get,
\[X=\dfrac{-\left( -7 \right)+\sqrt{121}}{2\times \left( 6 \right)}\]
And, \[X=\dfrac{-\left( -7 \right)-\sqrt{121}}{2\times \left( 6 \right)}\]
We know that \[\sqrt{121}=11\], so we get,
\[X=\dfrac{7+11}{12}=\dfrac{18}{12}=\dfrac{3}{2}\]
And \[X=\dfrac{7-11}{12}=\dfrac{-4}{12}=\dfrac{-1}{3}\]
So, we get the roots of the equation \[6{{x}^{2}}-7x-3=0\] as \[\dfrac{3}{2}\] and \[\dfrac{-1}{3}\].
Note:
Students can also find the zeroes of the quadratic equation \[6{{x}^{2}}-7x-3\] as follows:
\[6{{x}^{2}}-7x-3=0\]
Here, we can write 7x = 9x – 2x, we get,
\[6{{x}^{2}}-\left( 9x-2x \right)-3=0\]
Or, \[6{{x}^{2}}-9x+2x-3=0\]
We can also write the above equation as,
\[3x\left( 2x-3 \right)+\left( 2x-3 \right)=0\]
By taking (2x – 3) common, we get,
\[\left( 2x-3 \right)\left( 3x+1 \right)=0\]
So, we get \[\left( 2x-3 \right)=0\] and \[\left( 3x+1 \right)=0\]
\[x=\dfrac{3}{2}\text{ and }x=\dfrac{-1}{3}\]
So, we get the roots of the quadratic equation as \[\dfrac{3}{2}\] and \[\dfrac{-1}{3}\].
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
Mark and label the given geoinformation on the outline class 11 social science CBSE
When people say No pun intended what does that mea class 8 english CBSE
Name the states which share their boundary with Indias class 9 social science CBSE
Give an account of the Northern Plains of India class 9 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
In Indian rupees 1 trillion is equal to how many c class 8 maths CBSE
Which are the Top 10 Largest Countries of the World?
How do you graph the function fx 4x class 9 maths CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
Difference Between Plant Cell and Animal Cell
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Why is there a time difference of about 5 hours between class 10 social science CBSE