To determine which of the equations and their corresponding solutions are matched correctly, let's solve each equation one by one.
-
Equation: \(-x^2 - 64 = 0\)
Rearranging gives: \[ -x^2 = 64 \] \[ x^2 = -64 \]
This equation has no real solutions (since the square of a number cannot be negative). Therefore, the provided solution \( x = -8, 8 \) is incorrect.
-
Equation: \(-x^2 - x - 70 = 0\)
Rearranging gives: \[ -x^2 - x = 70 \] \[ x^2 + x + 70 = 0\]
Here, we can apply the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\): \[a = 1, b = 1, c = 70\] \[b^2 - 4ac = 1^2 - 4 \cdot 1 \cdot 70 = 1 - 280 = -279\]
Again, this equation has no real solutions. Thus, the provided solution \( x = -10, 7 \) is also incorrect.
-
Equation: \(4x^2 - 3x^2 + 10x = -25\)
Rearranging gives: \[ x^2 + 10x + 25 = 0 \] This can be factored as: \[(x + 5)^2 = 0\] Thus, \(x = -5\) (a double root).
The provided solution \(x = -5\) is correct.
-
Equation: \(x^2 + 6x = 112\)
Rearranging gives: \[x^2 + 6x - 112 = 0\]
Applying the quadratic formula: \[a = 1, b = 6, c = -112\] \[b^2 - 4ac = 6^2 - 4 \cdot 1 \cdot (-112) = 36 + 448 = 484\] \[\sqrt{484} = 22\] Thus: \[x = \frac{-6 \pm 22}{2}\] This gives us: \[x = \frac{16}{2} = 8, \quad x = \frac{-28}{2} = -14\]
Therefore, the solution \(x = -8, 14\) is incorrect.
In summary, the correctly matched equation and solution are:
- \(4x^2 - 3x^2 + 10x = -25\), \(x = -5\)
Therefore, the correct response is: 4x2−3x2+10x=−25 , x=−5