(a-3) x² + 4 (a-3) x+4 =0 has equal roots. Find the value of a
Answers
Answer :
a = 19
Note:
★ The possible values of the variable which satisfy the equation are called its roots or solutions .
★ A quadratic equation can have atmost two roots .
★ The general form of a quadratic equation is given as ; Ax² + Bx + C = 0
★ If α and ß are the roots of the quadratic equation Ax² + Bx + C = 0 , then ;
• Sum of roots , (α + ß) = -B/A
• Product of roots , (αß) = C/A
★ If α and ß are the roots of a quadratic equation , then that quadratic equation is given as : k•[ x² - (α + ß)x + αß ] = 0 , k ≠ 0.
★ The discriminant , D of the quadratic equation Ax² + Bx + C = 0 is given by ;
D = B² - 4AC
★ If D = 0 , then the roots are real and equal .
★ If D > 0 , then the roots are real and distinct .
★ If D < 0 , then the roots are unreal (imaginary) .
Solution :
Here ,
The given quadratic equation is ;
(a-3)x² + 4(a-3)x + 4 = 0
Now ,
Comparing the given quadratic equation with the general quadratic equation Ax² + Bx + C = 0 , we have ;
A = (a - 3)
B = (a - 3)
C = 4
Now ,
The discriminant of the given quadratic equation will be given as ;
=> D = B² - 4AC
=> D = (a - 3)² - 4×(a - 3)×4
=> D = (a - 3)² - 16(a - 3)
=> D = (a - 3)(a - 3 - 16)
=> D = (a - 3)(a - 19)
Now ,
The given quadratic equation will have equal roots if its discriminant is zero .
Thus ,
=> D = 0
=> (a - 3)(a - 19) = 0
=> a = 3 , 19
Here ,
a = 3 is rejected value , because if a = 3 , the given equation will no more be a quadratic equation .
Thus ,
a = 19 is appropriate value .
Hence ,
Required value of a is 19 .
Answer:
a=4
Step-by-step explanation:
(a-3)x^2+4 (a-3)x+4=0
since the equation has equal roots
D=0
b^2-4ac=0
on comparing : a=(a-3),b=4×(a-3),c=4
b^2-4ac=0
=(4 (a-3) )^2-4 (a-3)4=0
=16 (a-3)^2-16 (a-3)=0
=16 (a-3)^2×(a-3-1)=0
=16 (a-3) (a-4)=0
=(a-3) (a-4) =0
a=3, a=4
so a=4 is the values of A such that quadratic equation has equal roots