Question

Expand and simplify each of the following.(b) (2c + 1) (c - 5) + (c - 4) (0 - 3)​

Answer 1

The value of (2c+1) (c-5)+(c-4)(0-3)​ is  $\alpha =\frac{(6+\sqrt{22 }) }{2}$  and  $\beta =\frac{(6-\sqrt{22 }) }{2}$

Explanation:

Given:

(2c + 1) (c - 5) + (c - 4) (0 - 3)​

Formula:

$\alpha =\frac{-b+\sqrt{b^{2}-4ac } }{2a}$

$\beta=\frac{-b-\sqrt{b^{2}-4ac } }{2a}$

Solution:

==> Multiply (2c + 1) (c - 5)

==> (2c) (c - 5) + 1(c - 5)

==> 2c²-10c+c-5

==> 2c²-9c-5

==> (2c + 1) (c - 5)=2c²-9c-5

==>Multiply (c - 4) (0 - 3)​

==>(c-4)(-3)

==> -3c+12

==>  (c - 4) (0 - 3)​=-3c+12

==> Add the values

==> 2c²-9c-5-3c+12

==> 2c²-9c-3c+12-5

==>  2c²-12c+7

==> This is the quadratic equation

The value of (2c + 1) (c - 5) + (c - 4) (0 - 3)​ is  2c²-12c+7

==>2c²-12c+7

==> a = 2

==> b = -12

==> c = 7

==> Apply the a,b and c values in the formula

==> $\alpha =\frac{-b+\sqrt{b^{2}-4ac } }{2a}$

==> $\alpha =\frac{-(-12)+\sqrt{(-12)^{2}-4(2)(7) } }{2(2)}$

==> $\alpha =\frac{12+\sqrt{144-4(14) } }{4}$

==> $\alpha =\frac{12+\sqrt{144-56 } }{4}$

==> $\alpha =\frac{12+\sqrt{88 } }{4}$

==> $\alpha =\frac{12+\sqrt{11\times\f2\times2\times2 } }{4}$

==> $\alpha =\frac{12+2\sqrt{22 } }{4}$

==> $\alpha =\frac{2(6+\sqrt{22 }) }{4}$

==> $\alpha =\frac{(6+\sqrt{22 }) }{2}$

==> To find another values ,

==> $\beta =\frac{-b-\sqrt{b^{2}-4ac } }{2a}$

==> $\beta =\frac{-(-12)-\sqrt{(-12)^{2}-4(2)(7) } }{2(2)}$

==> $\beta =\frac{12-\sqrt{144-4(14) } }{4}$

==> $\beta =\frac{12-\sqrt{144-56 } }{4}$

==> $\beta =\frac{12-\sqrt{88 } }{4}$

==> $\beta=\frac{12-\sqrt{11\times\f2\times2\times2 } }{4}$

==> $\beta =\frac{12-2\sqrt{22 } }{4}$

==> $\beta =\frac{2(6-\sqrt{22 }) }{4}$

==> $\beta =\frac{(6-\sqrt{22 }) }{2}$

The value of (2c+1) (c-5)+(c-4)(0-3)​ is  $\alpha =\frac{(6+\sqrt{22 }) }{2}$  and  $\beta =\frac{(6-\sqrt{22 }) }{2}$