Answer:
Part 1) In the procedure
Part 2) In the procedure
Step-by-step explanation:
Part 1) Observing the graph
The roots are x=-1/2 and x=4
Remember that, the roots are the x-values when the value of y is equal to zero (the x-intercepts)
That's why Sheldon immediately realizes Howard's solution is incorrect
Part 2) we have
[tex]y=2x^{2}-7x-4[/tex]
The formula to solve a quadratic equation of the form [tex]ax^{2} +bx+c=0[/tex] is equal to
[tex]x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}[/tex]
in this problem we have
equate the equation to zero
[tex]2x^{2}-7x-4=0[/tex]
so
[tex]a=2\\b=-7\\c=-4[/tex]
substitute
[tex]x=\frac{-(-7)(+/-)\sqrt{-7^{2}-4(2)(-4)}} {2(2)}[/tex]
[tex]x=\frac{7(+/-)\sqrt{49+32}} {4}[/tex]
Howard's error is in this step, is wrong with the sign of the number 7, is positive instead of negative
[tex]x=\frac{7(+/-)\sqrt{81}}{4}[/tex]
[tex]x=\frac{7(+/-)9}{4}[/tex]
[tex]x=\frac{7(+)9}{4}=4[/tex]
[tex]x=\frac{7(-)9}{4}=-1/2[/tex]
