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If the equation [tex][tex]$a^2 x^2 + 6ab x + ac + 8b^2 = 0$[/tex][/tex] has equal roots, prove that the roots of the equation [tex][tex]$ac(x + 1)^2 = 4b^2 x$[/tex][/tex] are also equal.

Sagot :

GinnyAnswer
Of course! Let’s solve the problem step-by-step:

### Step 1: Analyze the given quadratic equation

Given the quadratic equation:
[tex]\[ a^2 x^2 + 6 a b x + a c + 8 b^2 = 0 \][/tex]

For quadratic equations of the form [tex]\( Ax^2 + Bx + C = 0 \)[/tex] to have equal roots, the discriminant must be zero. The discriminant [tex]\( \Delta \)[/tex] is given by:
[tex]\[ \Delta = B^2 - 4AC \][/tex]

### Step 2: Identify coefficients

In our equation:
[tex]\[ a^2 x^2 + 6 a b x + a c + 8 b^2 = 0 \][/tex]

The coefficients are:
[tex]\[ A = a^2, \quad B = 6ab, \quad C = ac + 8b^2 \][/tex]

### Step 3: Compute the discriminant

The discriminant for our quadratic equation is:
[tex]\[ \Delta = B^2 - 4AC \][/tex]
Substituting the coefficients, we get:
[tex]\[ \Delta = (6ab)^2 - 4(a^2)(ac + 8b^2) \][/tex]
[tex]\[ \Delta = 36a^2b^2 - 4a^2(ac + 8b^2) \][/tex]
[tex]\[ \Delta = 36a^2b^2 - 4a^3c - 32a^2b^2 \][/tex]
[tex]\[ \Delta = 4a^2b^2 - 4a^3c \][/tex]
[tex]\[ \Delta = 4a^2(b^2 - ac) \][/tex]

For the roots to be equal, the discriminant must be zero:
[tex]\[ 4a^2(b^2 - ac) = 0 \][/tex]

Since [tex]\( a^2 \neq 0 \)[/tex], we can simplify this to:
[tex]\[ b^2 - ac = 0 \][/tex]
[tex]\[ b^2 = ac \][/tex]

### Step 4: Analyze the second equation

Given the second equation:
[tex]\[ ac(x+1)^2 = 4b^2 x \][/tex]

We want to show that this equation also has equal roots. Start by expanding and simplifying:
[tex]\[ ac(x^2 + 2x + 1) = 4b^2 x \][/tex]
[tex]\[ acx^2 + 2acx + ac = 4b^2x \][/tex]

Rearrange everything to one side:
[tex]\[ acx^2 + 2acx + ac - 4b^2x = 0 \][/tex]
[tex]\[ acx^2 + (2ac - 4b^2)x + ac = 0 \][/tex]

### Step 5: Compute the discriminant of the new equation

Identify the coefficients again:
[tex]\[ A' = ac, \quad B' = 2ac - 4b^2, \quad C' = ac \][/tex]

Compute the discriminant [tex]\( \Delta' \)[/tex]:
[tex]\[ \Delta' = (B')^2 - 4A'C' \][/tex]
[tex]\[ \Delta' = (2ac - 4b^2)^2 - 4(ac)(ac) \][/tex]
[tex]\[ \Delta' = (2ac - 4b^2)^2 - 4a^2c^2 \][/tex]
[tex]\[ \Delta' = 4( ac - 2b^2)^2 - 4a^2c^2 \][/tex]
[tex]\[ \Delta' = 4 \left( ac - 2b^2 \right)^2 - 4a^2c^2 \][/tex]

Since we know from earlier that [tex]\( b^2 = ac \)[/tex], substitute [tex]\( b^2 \)[/tex] back into the discriminant:
[tex]\[ \Delta' = 4 \left( ac - 2(ac) \right)^2 - 4a^2c^2 \][/tex]
[tex]\[ \Delta' = 4 \left( -ac \right)^2 - 4a^2c^2 \][/tex]
[tex]\[ \Delta' = 4a^2c^2 - 4a^2c^2 \][/tex]
[tex]\[ \Delta' = 0 \][/tex]

### Conclusion

The discriminant of the second equation is zero, indicating that it also has equal roots. So we have proven that if the equation [tex]\( a^2 x^2 + 6 a b x + a c + 8 b^2 = 0 \)[/tex] has equal roots, the equation [tex]\( a c(x+1)^2 = 4 b^2 x \)[/tex] also has equal roots.
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