Question

Plz help in solving these 2 questions with steps and explanation!!

Answer 1

Explanation:

x^2 - 7x + 10 = 0 can be factored as follows: (x - 5)(x - 2). Note that -5x -2x combine to -7x, the middle term of this quadratic, and that (-5)(-2) = +10, the constant term. Setting each of these factors = to 0 separately, we get:

x = 5 and x = 2.

x^2 - 2x = 20 should be rewritten in standard form for a quadratic equation before you attempt to solve it: x^2 - 2x - 20 = 0. This quadratic is not so easily factored as was the previous one. Let's use the quadratic formula:

-b ± √(b²-4ac)

x = --------------------

2a

Here, a = 1, b = -2 and c = -20, so the discriminant b²-4ac = (-2)^2 - 4(1)(-20), or 4 + 80, or 84. 84 has only one perfect square factor: 4·21. Because the discriminant is +, we know that this equation has two real, unequal roots.

They are:

-(-2) ± √(4·21) 2 ± 2√21

x = ---------------------- = ----------------- = 1 ± √21

2(1) 2

Answer 2

`\large\boxed{Q1:\ x=2\ or\ x=5}\\\boxed{Q2:\ x=1-√(21)\ or\ x=1+√(21)}`

Explanation:

`\text{Use the quadratic formula:}\\\\ax^2+bx+c=0\\\\\text{If}\ b^2-4ac<0,\ \text{then the equation has}\ \bold{no\ solution}\\\\\text{If}\ b^2-4ac=0,\ \text{then the equation has one solution}\ x=(-b)/(2a)\\\\\text{If}\ b^2-4ac>0,\ \text{then the equation has two solutions}\ x=(-b\pm√(b^2-4ac))/(2a)\\\\==========================================`

`\bold{Q1}\\\\x^2-7x+10=0\\\\a=1,\ b=-7,\ c=10\\\\b^2-4ac=(-7)^2-4(1)(10)=49-40=9>0\\\\√(b^2-4ac)=\sqrt9=3\\\\x_1=(-(-7)-3)/(2(1))=(7-3)/(2)=(4)/(2)=2\\\\x_2=(-(-7)+3)/(2(1))=(7+3)/(2)=(10)/(2)=5\\\\========================================`

`\bold{Q2}\\x^2-2x=20\qquad\text{subtract 20 from both sides}\\\\x^2-2x-20=0\\\\a=1,\ b=-2,\ c=-20\\\\b^2-4ac=(-2)^2-4(1)(-20)=4+80=84>0\\\\√(b^2-4ac)=√(84)=√(4\cdot21)=\sqrt4\cdot√(21)=2√(21)\\\\x_1=(-(-2)-2√(21))/(2(1))=(2-2√(21))/(2)=1-√(21)\\\\x_2=(-(-2)+2√(21))/(2(1))=(2+2√(21))/(2)=1+√(21)`