Answer:
This implies that
x2+2ax=4x−4a−13
or
x2+2ax−4x+4a+13=0
or
x2+(2a−4)x+(4a+13)=0
Since the equation has just one solution instead of the usual two distinct solutions, then the two solutions must be same i.e. discriminant = 0.
Hence we get that
(2a−4)2=4⋅1⋅(4a+13)
or
4a2−16a+16=16a+52
or
4a2−32a−36=0
or
a2−8a−9=0
or
(a−9)(a+1)=0
So the values of a are −1 and 9.
Step-by-step explanation:
Answer:
[tex]\frac{492}{65}[/tex]
Step-by-step explanation:
Using the trigonometric identity
sin²A + cos²A = 1 , then
sinA = [tex]\sqrt{1-cos^2A}[/tex] thus
sinA = [tex]\sqrt{1-(5/13)^2}[/tex]
= [tex]\sqrt{1-\frac{25}{169} }[/tex]
= [tex]\sqrt{\frac{144}{169} }[/tex] = [tex]\frac{12}{13}[/tex] , then
tanA = [tex]\frac{sinA}{cosA}[/tex] = [tex]\frac{\frac{12}{13} }{\frac{5}{13} }[/tex] = [tex]\frac{12}{5}[/tex]
Thus
2tanA + 3sinA
= 2 × [tex]\frac{12}{5}[/tex] + ( 3 × [tex]\frac{12}{13}[/tex] )
= [tex]\frac{24}{5}[/tex] + [tex]\frac{36}{13}[/tex]
= [tex]\frac{312}{65}[/tex] + [tex]\frac{180}{65}[/tex]
= [tex]\frac{492}{65}[/tex]
