Math / Algebra

Question $\begin{array}{l} \text { 2. } y=x^{2}-6 x+5 \\ y=12 \\ \begin{array}{l} 12=x^{2}-6 x+5 \\ -12 \quad-12 \end{array} \\ x^{2}-6 x-7=0 \end{array}$

Studdy Solution

STEP 1

1. The problem involves solving a quadratic equation.
2. The equation is given in the form $y = x^2 - 6x + 5$ and we need to find $x$ when $y = 12$ .
3. The quadratic equation can be solved by factoring, completing the square, or using the quadratic formula.

STEP 2

1. Substitute the given value of $y$ into the equation.
2. Rearrange the equation to standard quadratic form.
3. Solve the quadratic equation using an appropriate method.
4. Verify the solutions.

STEP 3

Substitute $y = 12$ into the equation $y = x^2 - 6x + 5$ :
$12 = x^2 - 6x + 5$

STEP 4

Rearrange the equation to standard quadratic form by moving all terms to one side:
$12 = x^2 - 6x + 5$ Subtract 12 from both sides:
$x^2 - 6x + 5 - 12 = 0$ $x^2 - 6x - 7 = 0$

STEP 5

Solve the quadratic equation $x^2 - 6x - 7 = 0$ . We will use the quadratic formula, which is given by:
$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
For the equation $x^2 - 6x - 7 = 0$ , identify $a = 1$ , $b = -6$ , and $c = -7$ .

STEP 6

Calculate the discriminant $b^2 - 4ac$ :
$b^2 - 4ac = (-6)^2 - 4(1)(-7)$ $= 36 + 28$ $= 64$

STEP 7

Substitute the values into the quadratic formula:
$x = \frac{-(-6) \pm \sqrt{64}}{2(1)}$ $x = \frac{6 \pm 8}{2}$

STEP 8

Calculate the two possible values for $x$ :
$x = \frac{6 + 8}{2} = \frac{14}{2} = 7$ $x = \frac{6 - 8}{2} = \frac{-2}{2} = -1$

STEP 9

Verify the solutions by substituting back into the original equation $y = x^2 - 6x + 5$ to ensure they satisfy $y = 12$ .
For $x = 7$ :
$y = 7^2 - 6(7) + 5 = 49 - 42 + 5 = 12$
For $x = -1$ :
$y = (-1)^2 - 6(-1) + 5 = 1 + 6 + 5 = 12$
Both solutions are verified.
The solutions for $x$ are:
$\boxed{7 \text{ and } -1}$

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Question $\begin{array}{l} \text { 2. } y=x^{2}-6 x+5 \\ y=12 \\ \begin{array}{l} 12=x^{2}-6 x+5 \\ -12 \quad-12 \end{array} \\ x^{2}-6 x-7=0 \end{array}$

Studdy Solution

STEP 1

1. The problem involves solving a quadratic equation.
2. The equation is given in the form $y = x^2 - 6x + 5$ and we need to find $x$ when $y = 12$ .
3. The quadratic equation can be solved by factoring, completing the square, or using the quadratic formula.

STEP 2

1. Substitute the given value of $y$ into the equation.
2. Rearrange the equation to standard quadratic form.
3. Solve the quadratic equation using an appropriate method.
4. Verify the solutions.

STEP 3

Substitute $y = 12$ into the equation $y = x^2 - 6x + 5$ :
$12 = x^2 - 6x + 5$

STEP 4

Rearrange the equation to standard quadratic form by moving all terms to one side:
$12 = x^2 - 6x + 5$ Subtract 12 from both sides:
$x^2 - 6x + 5 - 12 = 0$ $x^2 - 6x - 7 = 0$

STEP 5

Solve the quadratic equation $x^2 - 6x - 7 = 0$ . We will use the quadratic formula, which is given by:
$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
For the equation $x^2 - 6x - 7 = 0$ , identify $a = 1$ , $b = -6$ , and $c = -7$ .

STEP 6

Calculate the discriminant $b^2 - 4ac$ :
$b^2 - 4ac = (-6)^2 - 4(1)(-7)$ $= 36 + 28$ $= 64$

STEP 7

Substitute the values into the quadratic formula:
$x = \frac{-(-6) \pm \sqrt{64}}{2(1)}$ $x = \frac{6 \pm 8}{2}$

STEP 8

Calculate the two possible values for $x$ :
$x = \frac{6 + 8}{2} = \frac{14}{2} = 7$ $x = \frac{6 - 8}{2} = \frac{-2}{2} = -1$

STEP 9

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Question 2. y=x2−6x+5y=1212=x2−6x+5−12−12x2−6x−7=0\begin{array}{l} \text { 2. } y=x^{2}-6 x+5 \\ y=12 \\ \begin{array}{l} 12=x^{2}-6 x+5 \\ -12 \quad-12 \end{array} \\ x^{2}-6 x-7=0 \end{array} 2. y=x2−6x+5y=1212=x2−6x+5−12−12​x2−6x−7=0​

Studdy Solution

STEP 1

1. The problem involves solving a quadratic equation.2. The equation is given in the form y=x2−6x+5 y = x^2 - 6x + 5 y=x2−6x+5 and we need to find x x x when y=12 y = 12 y=12.3. The quadratic equation can be solved by factoring, completing the square, or using the quadratic formula.

STEP 2

1. Substitute the given value of y y y into the equation.2. Rearrange the equation to standard quadratic form.3. Solve the quadratic equation using an appropriate method.4. Verify the solutions.

STEP 3

Substitute y=12 y = 12 y=12 into the equation y=x2−6x+5 y = x^2 - 6x + 5 y=x2−6x+5:12=x2−6x+5 12 = x^2 - 6x + 5 12=x2−6x+5

STEP 4

Rearrange the equation to standard quadratic form by moving all terms to one side:12=x2−6x+5 12 = x^2 - 6x + 5 12=x2−6x+5 Subtract 12 from both sides:x2−6x+5−12=0 x^2 - 6x + 5 - 12 = 0 x2−6x+5−12=0 x2−6x−7=0 x^2 - 6x - 7 = 0 x2−6x−7=0

STEP 5

STEP 6

Calculate the discriminant b2−4ac b^2 - 4ac b2−4ac:b2−4ac=(−6)2−4(1)(−7) b^2 - 4ac = (-6)^2 - 4(1)(-7) b2−4ac=(−6)2−4(1)(−7) =36+28 = 36 + 28 =36+28 =64 = 64 =64

STEP 7

Substitute the values into the quadratic formula:x=−(−6)±642(1) x = \frac{-(-6) \pm \sqrt{64}}{2(1)} x=2(1)−(−6)±64​​ x=6±82 x = \frac{6 \pm 8}{2} x=26±8​

STEP 8

Calculate the two possible values for x x x:x=6+82=142=7 x = \frac{6 + 8}{2} = \frac{14}{2} = 7 x=26+8​=214​=7 x=6−82=−22=−1 x = \frac{6 - 8}{2} = \frac{-2}{2} = -1 x=26−8​=2−2​=−1

STEP 9

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Question $\begin{array}{l} \text { 2. } y=x^{2}-6 x+5 \\ y=12 \\ \begin{array}{l} 12=x^{2}-6 x+5 \\ -12 \quad-12 \end{array} \\ x^{2}-6 x-7=0 \end{array}$

1. The problem involves solving a quadratic equation.
2. The equation is given in the form $y = x^2 - 6x + 5$ and we need to find $x$ when $y = 12$ .
3. The quadratic equation can be solved by factoring, completing the square, or using the quadratic formula.

1. Substitute the given value of $y$ into the equation.
2. Rearrange the equation to standard quadratic form.
3. Solve the quadratic equation using an appropriate method.
4. Verify the solutions.

Substitute $y = 12$ into the equation $y = x^2 - 6x + 5$ :
$12 = x^2 - 6x + 5$

Rearrange the equation to standard quadratic form by moving all terms to one side:
$12 = x^2 - 6x + 5$ Subtract 12 from both sides:
$x^2 - 6x + 5 - 12 = 0$ $x^2 - 6x - 7 = 0$

Calculate the discriminant $b^2 - 4ac$ :
$b^2 - 4ac = (-6)^2 - 4(1)(-7)$ $= 36 + 28$ $= 64$

Substitute the values into the quadratic formula:
$x = \frac{-(-6) \pm \sqrt{64}}{2(1)}$ $x = \frac{6 \pm 8}{2}$

Calculate the two possible values for $x$ :
$x = \frac{6 + 8}{2} = \frac{14}{2} = 7$ $x = \frac{6 - 8}{2} = \frac{-2}{2} = -1$