
How do you solve \[2{{x}^{2}}-10x+7=0\] using the quadratic formula?
Answer
452.1k+ views
Hint: This type of problem is based on the concept of solving quadratic equations. First, we have to consider the whole function and then use quadratic formula to find the value of ‘x’, that is, \[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]. Here, a=2, b=-10 and c=7. We need to find the values of the given equation by making necessary calculations. And then solve x by making some adjustments to the given inequality.
Complete step by step solution:
According to the question, we are asked to solve the given equation \[2{{x}^{2}}-10x+7=0\].
We have been given the equation is \[2{{x}^{2}}-10x+7=0\]. -----(1)
We know that for a quadratic equation \[a{{x}^{2}}+bx+c=0\],
\[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\].
Here, by comparing with equation (1), we get,
a=2, b=-10 and c=-7.
And also \[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\].
Therefore,
\[x=\dfrac{-\left( -10 \right)\pm \sqrt{{{\left( -10 \right)}^{2}}-4\left( 2 \right)\left( 7 \right)}}{2\left( 2 \right)}\]
On further simplifications, we get,
\[x=\dfrac{10\pm \sqrt{{{\left( -10 \right)}^{2}}-4\left( 2 \right)\left( 7 \right)}}{4}\]
We know that \[{{\left( -10 \right)}^{2}}=100\].
Using this to find the value of x, we get
\[\Rightarrow x=\dfrac{10\pm \sqrt{100-8\left( 7 \right)}}{4}\]
\[\Rightarrow x=\dfrac{10\pm \sqrt{100-56}}{4}\]
\[\Rightarrow x=\dfrac{10\pm \sqrt{44}}{4}\]
We know that \[\sqrt{44}\] can be written as \[\sqrt{44}=\sqrt{4\times 11}\].
Using the property \[\sqrt{ab}=\sqrt{a}\cdot \sqrt{b}\], we get
\[\sqrt{44}=\sqrt{4}\cdot \sqrt{11}\]
Since, \[\sqrt{4}=2\], we get
\[\sqrt{44}=2\sqrt{11}\].
We get,
\[\Rightarrow x=\dfrac{10\pm 2\sqrt{11}}{4}\]
We find that 2 are common in both the terms of the numerator. Taking 2 common from the numerator, we get
\[\Rightarrow x=\dfrac{2\left( 5\pm \sqrt{11} \right)}{4}\]
We can express the value of x as
\[x=\dfrac{2\left( 5\pm \sqrt{11} \right)}{2\times 2}\]
We find that 2 are common in both the numerator and denominator. Cancelling 2, we get
\[x=\dfrac{5\pm \sqrt{11}}{2}\]
Therefore, the values of ‘x’ are
\[\dfrac{5+\sqrt{11}}{2}\] and \[\dfrac{5-\sqrt{11}}{2}\].
Hence, the values of ‘u’ in the quadratic equation \[2{{x}^{2}}-10x+7=0\] are \[\dfrac{5+\sqrt{11}}{2}\] and \[\dfrac{5-\sqrt{11}}{2}\]
Note: Whenever you get this type of problem, we should always try to make the necessary changes in the given equation to get the final solution of the equation which will be the required answer. We should avoid calculation mistakes based on sign conventions. We should always make some necessary calculations to obtain zero in the right-hand side of the equation for easy calculations. Also this question can be solved by factoring the given equation.
Complete step by step solution:
According to the question, we are asked to solve the given equation \[2{{x}^{2}}-10x+7=0\].
We have been given the equation is \[2{{x}^{2}}-10x+7=0\]. -----(1)
We know that for a quadratic equation \[a{{x}^{2}}+bx+c=0\],
\[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\].
Here, by comparing with equation (1), we get,
a=2, b=-10 and c=-7.
And also \[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\].
Therefore,
\[x=\dfrac{-\left( -10 \right)\pm \sqrt{{{\left( -10 \right)}^{2}}-4\left( 2 \right)\left( 7 \right)}}{2\left( 2 \right)}\]
On further simplifications, we get,
\[x=\dfrac{10\pm \sqrt{{{\left( -10 \right)}^{2}}-4\left( 2 \right)\left( 7 \right)}}{4}\]
We know that \[{{\left( -10 \right)}^{2}}=100\].
Using this to find the value of x, we get
\[\Rightarrow x=\dfrac{10\pm \sqrt{100-8\left( 7 \right)}}{4}\]
\[\Rightarrow x=\dfrac{10\pm \sqrt{100-56}}{4}\]
\[\Rightarrow x=\dfrac{10\pm \sqrt{44}}{4}\]
We know that \[\sqrt{44}\] can be written as \[\sqrt{44}=\sqrt{4\times 11}\].
Using the property \[\sqrt{ab}=\sqrt{a}\cdot \sqrt{b}\], we get
\[\sqrt{44}=\sqrt{4}\cdot \sqrt{11}\]
Since, \[\sqrt{4}=2\], we get
\[\sqrt{44}=2\sqrt{11}\].
We get,
\[\Rightarrow x=\dfrac{10\pm 2\sqrt{11}}{4}\]
We find that 2 are common in both the terms of the numerator. Taking 2 common from the numerator, we get
\[\Rightarrow x=\dfrac{2\left( 5\pm \sqrt{11} \right)}{4}\]
We can express the value of x as
\[x=\dfrac{2\left( 5\pm \sqrt{11} \right)}{2\times 2}\]
We find that 2 are common in both the numerator and denominator. Cancelling 2, we get
\[x=\dfrac{5\pm \sqrt{11}}{2}\]
Therefore, the values of ‘x’ are
\[\dfrac{5+\sqrt{11}}{2}\] and \[\dfrac{5-\sqrt{11}}{2}\].
Hence, the values of ‘u’ in the quadratic equation \[2{{x}^{2}}-10x+7=0\] are \[\dfrac{5+\sqrt{11}}{2}\] and \[\dfrac{5-\sqrt{11}}{2}\]
Note: Whenever you get this type of problem, we should always try to make the necessary changes in the given equation to get the final solution of the equation which will be the required answer. We should avoid calculation mistakes based on sign conventions. We should always make some necessary calculations to obtain zero in the right-hand side of the equation for easy calculations. Also this question can be solved by factoring the given equation.
Recently Updated Pages
Master Class 10 General Knowledge: Engaging Questions & Answers for Success

Master Class 10 Computer Science: Engaging Questions & Answers for Success

Master Class 10 Science: Engaging Questions & Answers for Success

Master Class 10 Social Science: Engaging Questions & Answers for Success

Master Class 10 Maths: Engaging Questions & Answers for Success

Master Class 10 English: Engaging Questions & Answers for Success

Trending doubts
Truly whole mankind is one was declared by the Kannada class 10 social science CBSE

Explain the three major features of the shiwaliks class 10 social science CBSE

Distinguish between the reserved forests and protected class 10 biology CBSE

What are the public facilities provided by the government? Also explain each facility

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

Question An example of homologous organs is a Our arm class 10 biology CBSE
