Question

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Answer 1

Answer:

Part 3)

a) $x=0$

b) $x=0$ and $x=\frac{3}{4}$

c) $x=-3$ and $x=-2$

d) $n=1$ and $n=4$

e) $x=-1$  and $x=\frac{2}{3}$

f) $x=-1$  and $x=\frac{1}{2}$

Part  4)$x=10\ cm$

Step-by-step explanation:

Part 3) Solve the following quadratic equations.

case a) we have

$4x^2=0$

Divide by 4 both sides

$x^2=0$

take square root both sides

$x=0$

case b) we have

$4x^2-3x=0$

Factor x

$x(4x-3)=0$

so

One solution is

$x=0$

Second solution is

$(4x-3)=0$

$x=\frac{3}{4}$

case c) we have

$x^2+5x+6=0$

The formula to solve a quadratic equation of the form

$ax^{2} +bx+c=0$

is equal to

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

in this problem we have

$x^2+5x+6=0$

so

$a=1\\b=5\\c=6$

substitute in the formula

$x=\frac{-5\pm\sqrt{5^{2}-4(1)(6)}} {2(1)}$

$x=\frac{-5\pm\sqrt{1}} {2}$

$x=\frac{-5\pm1} {2}$

therefore

$x=\frac{-5+1} {2}=-2$

$x=\frac{-5-1} {2}=-3$

case d) we have

$n^2-5n+4=0$

The formula to solve a quadratic equation of the form

$an^{2} +bn+c=0$

is equal to

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

in this problem we have

$n^2-5n+4=0$

so

$a=1\\b=-5\\c=4$

substitute in the formula

$n=\frac{-(-5)\pm\sqrt{-5^{2}-4(1)(4)}} {2(1)}$

$n=\frac{5\pm\sqrt{9}} {2}$

$n=\frac{5\pm3} {2}$

therefore

$n=\frac{5+3} {2}=4$

$n=\frac{5-3} {2}=1$

case e) we have

$3x^2+x-2=0$

The formula to solve a quadratic equation of the form

$ax^{2} +bx+c=0$

is equal to

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

in this problem we have

$3x^2+x-2=0$

so

$a=3\\b=1\\c=-2$

substitute in the formula

$x=\frac{-1\pm\sqrt{1^{2}-4(3)(-2)}} {2(3)}$

$x=\frac{-1\pm\sqrt{25}} {6}$

$x=\frac{-1\pm5} {6}$

therefore

$x=\frac{-1+5} {6}=\frac{2}{3}$

$x=\frac{-1-5} {6}=-1$

case f) we have

$6x^2+3x-3=0$

The formula to solve a quadratic equation of the form

$ax^{2} +bx+c=0$

is equal to

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

in this problem we have

$6x^2+3x-3=0$

so

$a=6\\b=3\\c=-3$

substitute in the formula

$x=\frac{-3\pm\sqrt{3^{2}-4(6)(-3)}} {2(6)}$

$x=\frac{-3\pm\sqrt{81}} {12}$

$x=\frac{-3\pm9} {12}$

therefore

$x=\frac{-3+9}{12}=\frac{1}{2}$

$x=\frac{-3-9}{12}=-1$

Part 4) we know that

The area of rectangle is equal to

$A=LW$

we have

$A=66\ cm^2\\L=(x+1)\ cm\\W=(x-4)\ cm$

substitute

$66=(x+1)(x-4)$

solve for x

Apply distributive property right side

$66=x^2-4x+x-4\\x^2-3x-70=0$

solve the quadratic equation by formula

we have

$a=1\\b=-3\\c=-70$

substitute in the formula

$x=\frac{-(-3)\pm\sqrt{-3^{2}-4(1)(-70)}} {2(1)}$

$x=\frac{3\pm\sqrt{289}} {2}$

$x=\frac{3\pm17} {2}$

therefore

$x=\frac{3+17} {2}=10$

$x=\frac{3-17} {2}=-7$  ----> the value of x cannot be negative

so

The solution is x=10 cm

$L=(10+1)=11\ cm\\W=10-4=6\ cm$