Question

1. $x^{2}+9 = 3$
2. $7(x^{2}+10)=5x^{2}+80$
3. $6x^{2}+125 = 5(x^{2}+25)$
4. $2x^{2}-100=x^{2}-19$
5. $15=x^{2}-34$
6. $x^{2}-9 = 3x^{2}+41$

Explanation:

for 5:

Step1: Isolate \(x^2\)

\(x^2 = 3 - 9\)

Step2: Simplify right side

\(x^2 = -6\)

Step3: Real solution check

No real \(x\) (square ≥0)

for 6:

Step1: Expand left side

\(7x^2 + 70 = 5x^2 + 80\)

Step2: Subtract \(5x^2\)

\(2x^2 + 70 = 80\)

Step3: Subtract 70

\(2x^2 = 10\)

Step4: Divide by 2

\(x^2 = 5\)

Step5: Solve for \(x\)

\(x = \pm\sqrt{5}\)

for 7:

Step1: Expand right side

\(6x^2 + 125 = 5x^2 + 125\)

Step2: Subtract \(5x^2\)

\(x^2 + 125 = 125\)

Step3: Subtract 125

\(x^2 = 0\)

Step4: Solve for \(x\)

\(x = 0\)

for 8:

Step1: Subtract \(x^2\)

\(x^2 - 100 = -19\)

Step2: Add 100

\(x^2 = 81\)

Step3: Solve for \(x\)

\(x = \pm9\)

Answer:

1. No real solution
2. \(x = \pm\sqrt{5}\)
3. \(x = 0\)
4. \(x = \pm9\)
5. \(x = \pm7\)
6. No real solution