On the equation kx2 + (K + 2) x + K / 4 = 0 of X, there are two equal real roots

On the equation kx2 + (K + 2) x + K / 4 = 0 of X, there are two equal real roots

That is to say, △ = 0 and X & # 178; coefficient is not equal to 0
So K & # 178; + 4K + 4-K & # 178; = 0
So k = - 1
-x²+x-1/4=0
-(x-1/2)²=0
So X1 = x2 = 1 / 2

It is known that the equation kx2-4kx + K-5 = 0 about X has two equal real roots. Find the value of K and solve the equation

∵ the original equation has two equal real roots, ∵ K ≠ 0 and △ = 0, that is, 16k2-4k (K-5) = 0, ∵ k = − 53 or K = 0 (rounding), ∵ the original equation can be reduced to − 53x2 + 203x − 203 = 0, ∵ 53 (x2 − 4X + 4) = 0, ∵ (X-2) 2 = 0, ∵ X1 = x2 = 2

It is known that the equation kx2-4kx + K-5 = 0 about X has two equal real roots. Find the value of K and solve the equation

∵ the original equation has two equal real roots, ∵ K ≠ 0 and △ = 0, that is, 16k2-4k (K-5) = 0, ∵ k = − 53 or K = 0 (rounding), ∵ the original equation can be reduced to − 53x2 + 203x − 203 = 0, ∵ 53 (x2 − 4X + 4) = 0, ∵ (X-2) 2 = 0, ∵ X1 = x2 = 2