How do you solve $2 c^{2} - 7 c = - 5$ $2c^2 - 7c = -5$ ?

Question

How do you solve $2 c^{2} - 7 c = - 5$ $2c^2 - 7c = -5$ ?

2 Answers

Impact of this question

2960 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-04-05T03:38:54

1 and 5/2

Explanation:

$y = 2 c^{2} - 7 c + 5 = 0$ $y = 2c^2 - 7c + 5 = 0$ .
Since a + b + c = 0, use shortcut. The 2 real rots are: 1 and $\frac{c}{a} = \frac{5}{2}$ $c/a = 5/2$

Answer 2 · 2016-04-05T08:50:23

$x = \frac{5}{2}, 1$ $x=5/2,1$

Explanation:

$2 c^{2} - 7 c = - 5$ $color(blue)(2c^2-7c=-5$

Add $5$ $5$ both sides

$\to 2 c^{2} - 7 c + 5 = - 5 + 5$ $rarr2c^2-7c+5=-5+5$

$\to 2 c^{2} - 7 c + 5 = 0$ $rarr2c^2-7c+5=0$

Now,this is a trinomial.

So,

We can solve this by factoring or using Quadratic equation

$\approx \approx \approx \approx \approx \approx \approx \approx \approx \approx \approx \approx \approx \approx \approx \approx$ $~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$

Factoring

$2 c^{2} - 7 c + 5 = 0$ $color(purple)(2c^2-7c+5=0$

Factor the equation

If you have any problem with factoring trinomials,Watch this video:

$\to (2 x - 5) (x - 1) = 0$ $rarr(2x-5)(x-1)=0$

Now we can say that

$1) (x - 1) = 0$ $1)color(orange)((x-1)=0$

$2) (2 x - 5) = 0$ $2)color(indigo)((2x-5)=0$

Solve for both of the equations

$1) x - 1 = 0$ $1)color(orange)(x-1=0$

$\Rightarrow x = 1$ $color(green)(rArrx=1$

$2) 2 x - 5 = 0$ $2)color(indigo)(2x-5=0$

$\to 2 x = 5$ $rarr2x=5$

$\Rightarrow x = \frac{5}{2}$ $color(green)(rArrx=5/2$

$\approx \approx \approx \approx \approx \approx \approx \approx \approx \approx \approx \approx \approx \approx \approx$ $~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$

Using Quadratic formula

$2 c^{2} - 7 + 5 = 0$ $color(purple)(2c^2-7+5=0$

This is a Quadratic equation (in form $a x^{2} + b x + c = 0$ $ax^2+bx+c=0$ )

Quadratic formula

$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $color(brown)(x=(-b+-sqrt(b^2-4ac))/(2a)$

Remember that $a, b and c$ $a,bandc$ are the coefficients

So,

$a = 2, b = - 7, c = 5$ $color(violet)(a=2,b=-7,c=5$

$\to x = \frac{- (- 7) \pm \sqrt{- 7^{2} - 4 (2) (5)}}{2 (2)}$ $rarrx=(-(-7)+-sqrt(-7^2-4(2)(5)))/(2(2))$

$\to x = \frac{7 \pm \sqrt{49 - 4 (40)}}{4}$ $rarrx=(7+-sqrt(49-4(40)))/(4)$

$\to x = \frac{7 \pm \sqrt{49 - 40}}{4}$ $rarrx=(7+-sqrt(49-40))/(4)$

$\to x = \frac{7 \pm \sqrt{9}}{4}$ $rarrx=(7+-sqrt(9))/(4)$

$\to x = \frac{7 \pm 3}{4}$ $rarrx=(7+-3)/(4)$

Now we have two solutions

$1) x = \frac{7 + 3}{4} = \frac{10}{4} = \frac{5}{2}$ $1)color(orange)(x=(7+3)/(4)=10/4=5/2$

$2) x = \frac{7 - 3}{4} = \frac{4}{4} = 1$ $2)color(indigo)(x=(7-3)/4=4/4=1$

$\approx \approx \approx \approx \approx \approx \approx \approx \approx \approx \approx \approx \approx \approx \approx \approx$ $~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$

$∴ ¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯ ∣ ∣ ∣ x = \frac{5}{2}, 1 ∣ ∣ ∣ - ------- -$ $:.color(blue)(ul bar |x=5/2,1|$

Science

Math

Humanities

... and beyond

How do you solve $2 c^{2} - 7 c = - 5$ $2c^2 - 7c = -5$ ?

2 Answers

Explanation:

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you solve 2c^2 - 7c = -52c2−7c=−5 2c^2 - 7c = -5?

2 Answers

Explanation:

Explanation:

Related questions

Impact of this question

How do you solve $2 c^{2} - 7 c = - 5$ $2c^2 - 7c = -5$ ?