
Find the roots of ${{x}^{2}}+6x-8=0.$
Answer
507.3k+ views
Hint: We solve this question by using the formula method for finding out the roots. The formula for finding out the roots of a quadratic equation given by $a{{x}^{2}}+bx+c=0$ is given as $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}.$ By using this formula, we can compute the roots of a quadratic equation given in the question.
Complete step by step answer:
In order to solve this question, let us first check if the given equation in the question is in its standard form or not. A quadratic equation in its standard form is given by $a{{x}^{2}}+bx+c=0.$ The equation given in the question is,
$\Rightarrow {{x}^{2}}+6x-8=0$
This equation is given in the standard form. Comparing the two equations, we can see that $a=1,b=6,c=-8.$ We need these values to compute the roots of the given quadratic equation.
The formula for calculating the roots of a quadratic equation is given by,
$\Rightarrow x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
We use the above values of the coefficients a, b and c in the above equation to calculate the roots. Substituting them,
$\Rightarrow x=\dfrac{-6\pm \sqrt{{{6}^{2}}-4.1.-8}}{2.1}$
Multiplying the terms in the square root and squaring 6, and multiplying the terms in the denominator,
$\Rightarrow x=\dfrac{-6\pm \sqrt{36+32}}{2}$
Adding the terms in the square root,
$\Rightarrow x=\dfrac{-6\pm \sqrt{68}}{2}$
We know that 68 can be written as a product of 4 and 17.
$\Rightarrow x=\dfrac{-6\pm \sqrt{4\times 17}}{2}$
We can take the 4 outside the square root and it becomes 2,
$\Rightarrow x=\dfrac{-6\pm 2\sqrt{17}}{2}$
Dividing the first and second terms by 2,
$\Rightarrow x=-3\pm \sqrt{17}$
Hence, the two roots of the equation ${{x}^{2}}+6x-8=0$ is $-3+\sqrt{17}$ and $-3-\sqrt{17}.$
Note: We need to be careful while applying the formula as we need to substitute the right values from the given equation by comparing with the standard quadratic equation. If it is not in standard form, then we need to convert it to its standard form and then apply the formula. There are many ways to solve the quadratic equation for its roots such as by factoring, by formula, etc.
Complete step by step answer:
In order to solve this question, let us first check if the given equation in the question is in its standard form or not. A quadratic equation in its standard form is given by $a{{x}^{2}}+bx+c=0.$ The equation given in the question is,
$\Rightarrow {{x}^{2}}+6x-8=0$
This equation is given in the standard form. Comparing the two equations, we can see that $a=1,b=6,c=-8.$ We need these values to compute the roots of the given quadratic equation.
The formula for calculating the roots of a quadratic equation is given by,
$\Rightarrow x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
We use the above values of the coefficients a, b and c in the above equation to calculate the roots. Substituting them,
$\Rightarrow x=\dfrac{-6\pm \sqrt{{{6}^{2}}-4.1.-8}}{2.1}$
Multiplying the terms in the square root and squaring 6, and multiplying the terms in the denominator,
$\Rightarrow x=\dfrac{-6\pm \sqrt{36+32}}{2}$
Adding the terms in the square root,
$\Rightarrow x=\dfrac{-6\pm \sqrt{68}}{2}$
We know that 68 can be written as a product of 4 and 17.
$\Rightarrow x=\dfrac{-6\pm \sqrt{4\times 17}}{2}$
We can take the 4 outside the square root and it becomes 2,
$\Rightarrow x=\dfrac{-6\pm 2\sqrt{17}}{2}$
Dividing the first and second terms by 2,
$\Rightarrow x=-3\pm \sqrt{17}$
Hence, the two roots of the equation ${{x}^{2}}+6x-8=0$ is $-3+\sqrt{17}$ and $-3-\sqrt{17}.$
Note: We need to be careful while applying the formula as we need to substitute the right values from the given equation by comparing with the standard quadratic equation. If it is not in standard form, then we need to convert it to its standard form and then apply the formula. There are many ways to solve the quadratic equation for its roots such as by factoring, by formula, etc.
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