Answer
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Hint: In this question, we first need to apply the direct formula to get the values of x from the equation \[a{{x}^{2}}+bx+c=0,a\ne 0\]which is given by \[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]. Then from the obtained values of x we need to round them to one decimal to get the result.
Complete step-by-step answer:
As we already know that a quadratic polynomial \[f\left( x \right)\]when equated to zero is called a quadratic equation which is given by
\[a{{x}^{2}}+bx+c=0,a\ne 0\]
Direct formula: Quadratic equation \[a{{x}^{2}}+bx+c=0,\left( a\ne 0 \right)\]has two roots, given by
\[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]
Where, \[\Delta ={{b}^{2}}-4ac\]is called the discriminant of the equation.
Now, from the given quadratic equation in the question we have
\[5{{x}^{2}}+10x-3=0\]
Let us now compare this with the standard form of the quadratic equation to get the values of coefficients
\[a{{x}^{2}}+bx+c=0\]
Now, on comparison of the two equations we have
\[a=5,b=10,c=-3\]
Now, from the direct formula we have
\[\Rightarrow x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]
Let us now substitute the respective values
\[\Rightarrow x=\dfrac{-10\pm \sqrt{{{10}^{2}}-4\times 5\times \left( -3 \right)}}{2\times 5}\]
Now, this can be further written as
\[\Rightarrow x=\dfrac{-10\pm \sqrt{100+60}}{10}\]
Now, on further simplification we get,
\[\Rightarrow x=\dfrac{-10\pm \sqrt{160}}{10}\]
Now, this can be further written in the simplified form as
\[\Rightarrow x=\dfrac{-10\pm 4\sqrt{10}}{10}\]
Now, as we already know that the value of \[\sqrt{10}\]is 3.16
Let us now substitute this value in the above expression
\[\Rightarrow x=\dfrac{-10\pm 4\times 3.16}{10}\]
Now, on further simplification we get,
\[\Rightarrow x=\dfrac{-10\pm 12.64}{10}\]
Now, on further consideration this can be written as
\[\Rightarrow x=\dfrac{-10+12.64}{10},\dfrac{-10-12.64}{10}\]
Now, on simplifying them further we get,
\[\Rightarrow x=\dfrac{2.64}{10},\dfrac{-22.64}{10}\]
Now, this can also be written in the simplified form as
\[\Rightarrow x=0.264,-2.264\]
Let us now convert these to one decimal
\[\therefore x=0.3,-2.3\]
Note: Instead of using the direct formula we can also solve the equation using factorisation method. But, it would be difficult because we cannot check for all the square terms exactly. So, it is preferable to solve using the direct formula.
It is important to note that we should not neglect any of the terms while simplifying. It is also to be noted that we need to round the value of x to one decimal in which as the second decimal is greater than 5 the first decimal rises by 1 while rounding off.
Complete step-by-step answer:
As we already know that a quadratic polynomial \[f\left( x \right)\]when equated to zero is called a quadratic equation which is given by
\[a{{x}^{2}}+bx+c=0,a\ne 0\]
Direct formula: Quadratic equation \[a{{x}^{2}}+bx+c=0,\left( a\ne 0 \right)\]has two roots, given by
\[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]
Where, \[\Delta ={{b}^{2}}-4ac\]is called the discriminant of the equation.
Now, from the given quadratic equation in the question we have
\[5{{x}^{2}}+10x-3=0\]
Let us now compare this with the standard form of the quadratic equation to get the values of coefficients
\[a{{x}^{2}}+bx+c=0\]
Now, on comparison of the two equations we have
\[a=5,b=10,c=-3\]
Now, from the direct formula we have
\[\Rightarrow x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]
Let us now substitute the respective values
\[\Rightarrow x=\dfrac{-10\pm \sqrt{{{10}^{2}}-4\times 5\times \left( -3 \right)}}{2\times 5}\]
Now, this can be further written as
\[\Rightarrow x=\dfrac{-10\pm \sqrt{100+60}}{10}\]
Now, on further simplification we get,
\[\Rightarrow x=\dfrac{-10\pm \sqrt{160}}{10}\]
Now, this can be further written in the simplified form as
\[\Rightarrow x=\dfrac{-10\pm 4\sqrt{10}}{10}\]
Now, as we already know that the value of \[\sqrt{10}\]is 3.16
Let us now substitute this value in the above expression
\[\Rightarrow x=\dfrac{-10\pm 4\times 3.16}{10}\]
Now, on further simplification we get,
\[\Rightarrow x=\dfrac{-10\pm 12.64}{10}\]
Now, on further consideration this can be written as
\[\Rightarrow x=\dfrac{-10+12.64}{10},\dfrac{-10-12.64}{10}\]
Now, on simplifying them further we get,
\[\Rightarrow x=\dfrac{2.64}{10},\dfrac{-22.64}{10}\]
Now, this can also be written in the simplified form as
\[\Rightarrow x=0.264,-2.264\]
Let us now convert these to one decimal
\[\therefore x=0.3,-2.3\]
Note: Instead of using the direct formula we can also solve the equation using factorisation method. But, it would be difficult because we cannot check for all the square terms exactly. So, it is preferable to solve using the direct formula.
It is important to note that we should not neglect any of the terms while simplifying. It is also to be noted that we need to round the value of x to one decimal in which as the second decimal is greater than 5 the first decimal rises by 1 while rounding off.
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