Question

Check whether the following equations are quadratic or not.

(i) x+3/x=x2

(ii) 2x2 – 5x = x2 – 2x + 3
(iii) x2
1
= 5
x2
(iv) x2 – 3x - VX + 4 = 0
(V) x(2x + 3) = x + 2
(vi) (x - 2)² + 1 = 2x - 3
(vii) y(8y + 5) = y2 + 3
(viii) y(2y + 15) = 2(y2 + y + 8)​

Answer 1

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.

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