Question

solve by completing the square continued - 17) (x^{2}+10x + 3=0) (a = 1) (b = 10) (c = 3) -3 -3 (2) (a = 1) (b = 10) (\frac{10}{2}=5), (5^{2}=25) (x^{2}+10x=-3) -3 + 25 = 22 (x^{2}+10x + 25=-3 + 25) (sqrt{(x + 5)^{2}}=sqrt{22}) (x^{2}+10x + 25 = 22) ((x + 5)) (5xcdot2(10x)) (x + 5=pmsqrt{22}) -5 -5 (x=-5pmsqrt{22}) (x=-5pm2sqrt{7}) answer! 18) (x^{2}+12x + 5=0) -5 -5 (2) (a = 1) (b = 12) (x^{2}+12x=-5) (\frac{12}{2}=6), (6^{2}=36) (sqrt{(x + 6)^{2}}=sqrt{31}) (x^{2}+12x+36=-5 + 36) (x + 6=pmsqrt{31}) -6 -6 (x=-6pmsqrt{31}) 31 not factorable (x=-6pmsqrt{31})

Explanation:

Step1: Isolate the constant term for the first equation

For the quadratic equation \(x^{2}+10x + 3=0\), subtract 3 from both sides to get \(x^{2}+10x=-3\).

Step2: Complete the square

Take half of the coefficient of \(x\) (for \(x^{2}+10x\), half of 10 is 5), then square it (\(5^{2}=25\)). Add 25 to both sides of the equation: \(x^{2}+10x + 25=-3 + 25\), which simplifies to \((x + 5)^{2}=22\).

Step3: Solve for \(x\)

Take the square - root of both sides: \(x+5=\pm\sqrt{22}\), then \(x=-5\pm\sqrt{22}\).

For the second equation \(x^{2}+12x + 5=0\):

Step4: Isolate the constant term

Subtract 5 from both sides: \(x^{2}+12x=-5\).

Step5: Complete the square

Half of 12 is 6, and \(6^{2}=36\). Add 36 to both sides: \(x^{2}+12x + 36=-5 + 36\), so \((x + 6)^{2}=31\).

Step6: Solve for \(x\)

Take the square - root of both sides: \(x+6=\pm\sqrt{31}\), then \(x=-6\pm\sqrt{31}\).

Answer:

For \(x^{2}+10x + 3=0\), \(x=-5\pm\sqrt{22}\); for \(x^{2}+12x + 5=0\), \(x=-6\pm\sqrt{31}\)