Question

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

Answer 1

Answer:

Step-by-step 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

Answer:

$\large\boxed{Q1:\ x=2\ or\ x=5}\\\boxed{Q2:\ x=1-\sqrt{21}\ or\ x=1+\sqrt{21}}$

Step-by-step explanation:

$\text{Use the quadratic formula:}\\\\ax^2+bx+c=0\\\\\text{If}\ b^2-4ac0,\ \text{then the equation has two solutions}\ x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\==========================================$

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

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