Solve the following quadratic equation.
7x²+x-3=0
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Given Quadratic equation ➝ 7x² + x - 3 = 0
For finding the value of x, we apply the quadratic formula.
where:
[tex] \sf{x = \dfrac{ - b \pm \sqrt{ {b}^{2} - 4ac} }{2a} } \\ \\ [/tex]
[tex] \implies \sf{x = \dfrac{ - 1 \pm \sqrt{(1)^{2} - 4(7)( - 3)} }{2(7)} } \\ \\ [/tex]
[tex] \implies \sf{x = \dfrac{ - 1 \pm \sqrt{1 - ( - 84)} }{14} } \\ \\ [/tex]
[tex]\implies \sf{x = \dfrac{ - 1 \pm \sqrt{1 + 84} }{14} } \\ \\ [/tex]
[tex] \implies \sf{x = \dfrac{ - 1 \pm \sqrt{85} }{14} } \\ \\ [/tex]
[tex] \implies \sf{x = \dfrac{ - 1 \pm \sqrt{25 \times 3} }{14} } \\ \\ [/tex]
[tex] \implies \sf{x = \dfrac{ - 1 \pm \: 5 \sqrt{3} }{14} } \\ \\ [/tex]
Now there are 2 possible values of x that is:
[tex]\boxed{ \bf{x = \dfrac{5 \sqrt{3} - 1}{14} \quad or \quad x = \dfrac{ - 1 - 5 \sqrt{3} }{14} }}[/tex]
KNOW MORE:
For solving any given polynomial, there are 3 methods:
Given to Factorise:-
7x²+x-3=0
Formula used:-
[tex] \frac{ - b + - \sqrt{ {b}^{2} - 4ac} }{2a} [/tex]
Here,
a= 7
b= 1
c= -3
Applying the formula:-
[tex] \frac{ - 1 + - \sqrt{ {1}^{2} - 4(7)( - 3)} }{2(7)} [/tex]
[tex] \frac{ - 1 + - \sqrt{1 + 84} }{14} [/tex]
[tex] \frac{ - 1 + - \sqrt{85} }{14} [/tex]
[tex] = \frac{ - 1 + -9.219}{14} [/tex]
Hence,
The possible roots are:-
1st root=
[tex] \frac{ - 1 + 9.219}{14} [/tex]
[tex] = \frac{ 8.219}{14} [/tex]
[tex] = \frac{8.22}{14} [/tex]
=0.587
=0.59
2nd root-
[tex] \frac{ - 1 - 9.219}{14} [/tex]
[tex] = \frac{ - 10.219}{14} [/tex]
[tex] = \frac{ - 10.22}{14} [/tex]
= -0.73 (approx)
If we substitute the values of 1st root we get the value 0.0267 which is approximate 0 and if we substitute the 2nd root we get the value 0.0003 which is also approximate equals to 0 hence the given roots are:-
x= 0.59, -0.73
Note:If we don't round off the values we will get the value more close to zero but according to the rule we have to round off so I rounded off it.