Question

Use the quadratic formula to solve the equation.
2x^2 − 7x − 6 = 0
x= (smaller value)
x= (larger value)

Answer 1

Do you know the quadratic formula?

Answer 2

x=-b+and-sqrtb^2-4ac

a=2 b=7 c=6

Answer 3

over 2a

Answer 4

x=-bmoreorless sqrtb^2-4ac over 2a

Answer 5

You have the wrong b and c. You forgot your negatives:
a=2, b=-7, c = -6
Therefore
\[x=\frac{-(-7)\pm \sqrt{7^2 - 4(2)(-6)}}{2(2)}\]

Answer 6

\[2x^2 -7x-6=0\]
first determine the values of a, b, and c.
a= 2, b=-7 and c=-6
and substitute them to the quadratic formula:
\[x= (-b \pm \sqrt{b^2 -4ac}) \div2a \]

hence,
\[x= {7 \pm \sqrt{49-4(2)(-6)}} \div (2*2) \]
\[x=(7 \pm \sqrt{97}) \div 4\]

Answer 7

\[\left\{x\to \frac{1}{4} \left(7-\sqrt{97}\right),x\to \frac{1}{4} \left(7+\sqrt{97}\right)\right\}=\{x\to -0.712214,x\to 4.21221\} \]

Answer 8

x=4.212 is right but x=0.712214 its wrong

Answer 9

Replacing x in the expression:\[2x^2-7x-6 \]with\[\frac{1}{4} \left(7-\sqrt{97}\right) \]gives\[\frac{1}{8} \left(7-\sqrt{97}\right)^2-\frac{7}{4} \left(7-\sqrt{97}\right)-6 \]Expand the first term and simplify.\[\frac{73}{4}-\frac{7 \sqrt{97}}{4}-\frac{7}{4} \left(7-\sqrt{97}\right)-6 \]Expand the second term and combine with the third\[\frac{73}{4}-\frac{7 \sqrt{97}}{4}\left(-\frac{73}{4}+\frac{7 \sqrt{97}}{4}\right) \]The sum of the remaining terms is zero.