
How do you solve ${{x}^{2}}-3x-1=0$ using the quadratic formula?
Answer
490.5k+ views
Hint: Quadratic formula for the equation $a{{x}^{2}}+bx+c=0$ is given as $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$. Any quadratic equation can be compared with the general two degree equation $a{{x}^{2}}+bx+c=0$ to find the value of $a$, $b$ and $c$. After that direct substitution of the values in the quadratic formula gives the values of $x$ satisfying that quadratic equation.
Complete Step by Step Solution:
Compare the given equation ${{x}^{2}}-3x-1=0$ with $a{{x}^{2}}+bx+c=0$ and find the values of $a$, $b$ and $c$.
$\Rightarrow a{{x}^{2}}+bx+c=0$
Now, we can easily find the values of a, b and c, that means, we get,
$\Rightarrow 1\left( {{x}^{2}} \right)-3\left( x \right)-1=0$
$\therefore $ Values are $a=1$, $b=-3$ and $c=-1$.
Substitute the values of $a$, $b$ and $c$ in the expression ${{b}^{2}}-4ac$.
$\Rightarrow {{b}^{2}}-4ac={{\left( -3 \right)}^{2}}-4\left( 1 \right)\left( -1 \right)$
Solve the right side of the equation.
$\Rightarrow {{\left( -3 \right)}^{2}}-4\left( 1 \right)\left( -1 \right)=9+4$
$\Rightarrow {{\left( -3 \right)}^{2}}-4\left( 1 \right)\left( -1 \right)=13$
Substitute the values of $a$, $b$ and of the expression ${{b}^{2}}-4ac$ in the quadratic formula.
$\Rightarrow x=\dfrac{-\left( -3 \right)\pm \sqrt{\left( 13 \right)}}{2\left( 1 \right)}$
Simplify the right side of the equation further.
$\Rightarrow x=\dfrac{3\pm \sqrt{13}}{2}$
Consider the positive and negative sign to obtain two values of $x$.
$\Rightarrow x=\dfrac{3+\sqrt{13}}{2}$
$\Rightarrow x=\dfrac{3-\sqrt{13}}{2}$
Additional Information:
Expression ${{b}^{2}}-4ac$ is known as the determinant of the quadratic equation and denoted by $D$. This explains about the nature of the roots of the equation. If $D>0$ that implies roots will be real and distinct. If $D=0$ that implies roots will be real and equal. If $D<0$ that implies roots will be complex.
Note:
Compare the general quadratic equation and the given equation carefully because any error can lead to no solution, complex solution or solutions of some other equation. Find the value of expression ${{b}^{2}}-4ac$ before substituting the values of a, band c in the quadratic formula because it helps to visualize the solution as well as reduce the calculations.
Complete Step by Step Solution:
Compare the given equation ${{x}^{2}}-3x-1=0$ with $a{{x}^{2}}+bx+c=0$ and find the values of $a$, $b$ and $c$.
$\Rightarrow a{{x}^{2}}+bx+c=0$
Now, we can easily find the values of a, b and c, that means, we get,
$\Rightarrow 1\left( {{x}^{2}} \right)-3\left( x \right)-1=0$
$\therefore $ Values are $a=1$, $b=-3$ and $c=-1$.
Substitute the values of $a$, $b$ and $c$ in the expression ${{b}^{2}}-4ac$.
$\Rightarrow {{b}^{2}}-4ac={{\left( -3 \right)}^{2}}-4\left( 1 \right)\left( -1 \right)$
Solve the right side of the equation.
$\Rightarrow {{\left( -3 \right)}^{2}}-4\left( 1 \right)\left( -1 \right)=9+4$
$\Rightarrow {{\left( -3 \right)}^{2}}-4\left( 1 \right)\left( -1 \right)=13$
Substitute the values of $a$, $b$ and of the expression ${{b}^{2}}-4ac$ in the quadratic formula.
$\Rightarrow x=\dfrac{-\left( -3 \right)\pm \sqrt{\left( 13 \right)}}{2\left( 1 \right)}$
Simplify the right side of the equation further.
$\Rightarrow x=\dfrac{3\pm \sqrt{13}}{2}$
Consider the positive and negative sign to obtain two values of $x$.
$\Rightarrow x=\dfrac{3+\sqrt{13}}{2}$
$\Rightarrow x=\dfrac{3-\sqrt{13}}{2}$
Additional Information:
Expression ${{b}^{2}}-4ac$ is known as the determinant of the quadratic equation and denoted by $D$. This explains about the nature of the roots of the equation. If $D>0$ that implies roots will be real and distinct. If $D=0$ that implies roots will be real and equal. If $D<0$ that implies roots will be complex.
Note:
Compare the general quadratic equation and the given equation carefully because any error can lead to no solution, complex solution or solutions of some other equation. Find the value of expression ${{b}^{2}}-4ac$ before substituting the values of a, band c in the quadratic formula because it helps to visualize the solution as well as reduce the calculations.
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