How do you solve $7 z^{2} = 70 z - 147$ $7z^2=70z-147$ ?

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How do you solve $7 z^{2} = 70 z - 147$ $7z^2=70z-147$ ?

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Answer 1 · 2017-09-09T22:19:02

See a solution process belowL

Explanation:

First, write this equation in standard form by subtracting $70 z$ $color(red)(70z)$ and adding $147$ $color(blue)(147)$ to each side of the equation:

$7 z^{2} - 70 z + 147 = 70 z - 70 z - 147 + 147$ $7z^2 - color(red)(70z) + color(blue)(147) = 70z - color(red)(70z) - 147 + color(blue)(147)$

$7 z^{2} - 70 z + 147 = 0 - 0$ $7z^2 - 70z + 147 = 0 - 0$

$7 z^{2} - 70 z + 147 = 0$ $7z^2 - 70z + 147 = 0$

Next, divide each side of the equation by $7$ $color(red)(7)$ to reduce the coefficients:

$\frac{7 z^{2} - 70 z + 147}{7} = \frac{0}{7}$ $(7z^2 - 70z + 147)/color(red)(7) = 0/color(red)(7)$

$\frac{7 z^{2}}{7} - \frac{70 z}{7} + \frac{147}{7} = 0$ $(7z^2)/color(red)(7) - (70z)/color(red)(7) + 147/color(red)(7) = 0$

$1 z^{2} - 10 z + 21 = 0$ $1z^2 - 10z + 21 = 0$

$z^{2} - 10 z + 21 = 0$ $z^2 - 10z + 21 = 0$

Then we can factor the right side of the equation as:

$(z - 3) (z - 7) = 0$ $(z - 3)(z - 7) = 0$

Now, solve each term on the right side of the equation for $0$ $0$ :

Solution 1:

$z - 3 = 0$ $z - 3 = 0$

$z - 3 + 3 = 0 + 3$ $z - 3 + color(red)(3) = 0 + color(red)(3)$

$z - 0 = 3$ $z - 0 = 3$

$z = 3$ $z = 3$

Solution 2:

$z - 7 = 0$ $z - 7 = 0$

$z - 7 + 7 = 0 + 7$ $z - 7 + color(red)(7) = 0 + color(red)(7)$

$z - 0 = 7$ $z - 0 = 7$

$z = 7$ $z = 7$

**The Solutions Are: $z = 3$ $z = 3$ and $z = 7$ $z = 7$

We could also use the quadratic formula to solve this problem. The quadratic formula states:

For $a x^{2} + b x + c = 0$ $color(red)(a)x^2 + color(blue)(b)x + color(green)(c) = 0$ , the values of $x$ $x$ which are the solutions to the equation are given by:

$x = \frac{- b \pm \sqrt{b^{2} - (4 a c)}}{2 \cdot a}$ $x = (-color(blue)(b) +- sqrt(color(blue)(b)^2 - (4color(red)(a)color(green)(c))))/(2 * color(red)(a))$

Substituting:

$1$ $color(red)(1)$ for $a$ $color(red)(a)$

$- 10$ $color(blue)(-10)$ for $b$ $color(blue)(b)$

$21$ $color(green)(21)$ for $c$ $color(green)(c)$ gives:

$z = \frac{- (- 10) \pm \sqrt{{(- 10)}^{2} - (4 \cdot 1 \cdot 21)}}{2 \cdot 1}$ $z = (-color(blue)((-10)) +- sqrt(color(blue)((-10))^2 - (4 * color(red)(1) * color(green)(21))))/(2 * color(red)(1))$

$z = \frac{10 \pm \sqrt{100 - 84}}{2}$ $z = (10 +- sqrt(color(blue)(100 - 84)))/2$

$z = \frac{10 - \sqrt{16}}{2}$ $z = (10 - sqrt(16))/2$ and $z = \frac{10 + \sqrt{16}}{2}$ $z = (10 + sqrt(16))/2$

$z = \frac{10 - 4}{2}$ $z = (10 - 4)/2$ and $z = \frac{10 + 4}{2}$ $z = (10 + 4)/2$

$z = \frac{6}{2}$ $z = 6/2$ and $z = \frac{14}{2}$ $z = 14/2$

$z = 3$ $z = 3$ and $z = 7$ $z = 7$

Answer 2 · 2017-09-10T05:01:54

z = - 3 and z = -7

Explanation:

$7 z^{2} - 70 z + 147 = 0$ $7z^2 - 70z + 147 = 0$
Divide both sides by 7
$z^{2} - 10 z + 21 = 0$ $z^2 - 10z + 21 = 0$
Solve this quadratic equation by the new Transforming Method (Socratic, Google Search), when a = 1.
Find 2 real roots knowing the sum (-b = 10) and the product (c = 21).
They are: (-3) and (-7)

Answer 3 · 2017-11-24T21:12:43

$z = 3$ $z = 3$
$z = 7$ $z = 7$

Here's how to solve this problem rapidly in four easy steps.

Explanation:

Given
$7 z^{2} = 70 z - 147$ $7z^2=70z−147$
Solve for $z$ $z$

Step 1) Divide all the terms on both sides by 7
$z^{2} = 10 z - 21$ $z^2 = 10z - 21$

Step 2) Set the equation equal to 0.
Subtract $10 z$ $10z$ from both sides and add $21$ $21$ to both sides
$z^{2} - 10 z + 21 = 0$ $z^2 - 10z + 21 = 0$

Step 3) Factor
$(z - 3) (z - 7) = 0$ $(z - 3)( z - 7 ) = 0$

Step 4) Set each factor equal to 0 and solve for $z$ $z$

Don't write this step out if you can just predict the answers by looking at the factors $(z - 3) (z - 7) .$ $(z - 3)( z - 7 ).$

Just write down the answers and skip this step if you can.
................

Here's how to find the answers if you can't tell what they will be just by looking at $(z - 3) (z - 7) = 0$ $(z - 3)( z - 7 ) = 0$ (Step 3 above)

Set each factor equal to 0 and solve for $z$ $z$

$z - 3 = 0$ $z - 3 = 0$
$z = 3$ $z = 3$ $\leftarrow$ $larr$ One answer

$z - 7 = 0$ $z - 7 = 0$
$z = 7$ $z = 7$ $\leftarrow$ $larr$ The other answer

Answer:
$z = 3$ $z = 3$
$z = 7$ $z = 7$
..............................

Check

Sub each answer into the original equation to make sure it is still an equality,

Sub in 3 in the place of $z$ $z$
$7 z^{2} = 70 z - 147$ $7z^2=70z−147$
$7 (3^{2})$ $7(3^2)$ . . should equal . . $70 (3)$ $70(3)$ . . $- 147$ $- 147$
. $63$ $63$ . . . . should equal . . . $210$ $210$ . . . $- 147$ $- 147$
. $63$ $63$ . . . . . . does equal . . . . . . . . $63$ $63$
Check!

(Testing $7$ $7$ to see if it's the right answer too isn't worth the trouble.)

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How do you solve $7 z^{2} = 70 z - 147$ $7z^2=70z-147$ ?

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