2 9 9x²t 1 1 4x² - 33 then Prove P 5 20- -1 - 5 130 3r
Answers
Step-by-step explanation:
Step-by-step explanation:
Formulas for Solving Quadratic Equations
1. The roots of the quadratic equation: x = (-b ± √D)/2a, where D = b2 – 4ac
2. Nature of roots:
D > 0, roots are real and distinct (unequal)
D = 0, roots are real and equal (coincident)
D < 0, roots are imaginary and unequal
3. The roots (α + iβ), (α – iβ) are the conjugate pair of each other.
4. Sum and Product of roots: If α and β are the roots of a quadratic equation, then
S = α+β= -b/a = coefficient of x/coefficient of x2
P = αβ = c/a = constant term/coefficient of x2
5. Quadratic equation in the form of roots: x2 – (α+β)x + (αβ) = 0
6. The quadratic equations a1x2 + b1x + c1 = 0 and a2x2 + b2x + c2 = 0 have;
One common root if (b1c2 – b2c1)/(c1a2 – c2a1) = (c1a2 – c2a1)/(a1b2 – a2b1)
Both roots common if a1/a2 = b1/b2 = c1/c2
7. In quadratic equation ax2 + bx + c = 0 or [(x + b/2a)2 – D/4a2]
If a > 0, minimum value = 4ac – b2/4a at x = -b/2a.
If a < 0, maximum value 4ac – b2/4a at x= -b/2a.
8. If α, β, γ are roots of cubic equation ax3 + bx2 + cx + d = 0, then, α + β + γ = -b/a, αβ + βγ + λα = c/a, and αβγ = -d/a
Step-by-step explanation:
Step-by-step explanation:
\begin{gathered}\begin{gathered} \sf \red{\frac{x - 3}{x + 3} + \frac{x + 3}{x - 3} = 2 \frac{1}{2} } \\ \\ \implies \: \frac{(x - 3)^{2} + (x + 3)^{2} }{(x + 3)(x - 3)} = \frac{5}{2} \end{gathered} \end{gathered}
x+3
x−3
+
x−3
x+3
=2
2
1
⟹
(x+3)(x−3)
(x−3)
2
+(x+3)
2
=
2
5
\sf \implies \: \frac{ {x}^{2} - 6x + 9 + {x}^{2} + 6x + 9 }{ {x}^{2} - 9}⟹
x
2
−9
x
2
−6x+9+x
2
+6x+9
\begin{gathered}\begin{gathered} \sf \implies \: 2(2 {x}^{2} + 18) = 5( {x}^{2} - 9) \\ \sf \implies \: 4 {x}^{2} + 36 = 5x2 - 45 \\ \sf \implies {x}^{2} - 81 = 0 \\ \sf \implies \: {x}^{2} - {9}^{2} = 0\end{gathered} \end{gathered}
⟹2(2x
2
+18)=5(x
2
−9)
⟹4x
2
+36=5x2−45
⟹x
2
−81=0
⟹x
2
−9
2
=0
\begin{gathered}\begin{gathered}\sf \implies \: (x + 9)(x - 9) = 0 \\ \\ \sf \red{if \: x + 9 = 0 \: or \: x - 9 = 0} \\ \sf \red{then \: x \: = - 9 \: or \: x \: = 9}\end{gathered}\end{gathered}
⟹(x+9)(x−9)=0
ifx+9=0orx−9=0
thenx=−9orx=9
Formulas for Solving Quadratic Equations
1. The roots of the quadratic equation: x = (-b ± √D)/2a, where D = b2 – 4ac
2. Nature of roots:
D > 0, roots are real and distinct (unequal)
D = 0, roots are real and equal (coincident)
D < 0, roots are imaginary and unequal
3. The roots (α + iβ), (α – iβ) are the conjugate pair of each other.
4. Sum and Product of roots: If α and β are the roots of a quadratic equation, then
S = α+β= -b/a = coefficient of x/coefficient of x2
P = αβ = c/a = constant term/coefficient of x2
5. Quadratic equation in the form of roots: x2 – (α+β)x + (αβ) = 0
6. The quadratic equations a1x2 + b1x + c1 = 0 and a2x2 + b2x + c2 = 0 have;
One common root if (b1c2 – b2c1)/(c1a2 – c2a1) = (c1a2 – c2a1)/(a1b2 – a2b1)
Both roots common if a1/a2 = b1/b2 = c1/c2
7. In quadratic equation ax2 + bx + c = 0 or [(x + b/2a)2 – D/4a2]
If a > 0, minimum value = 4ac – b2/4a at x = -b/2a.
If a < 0, maximum value 4ac – b2/4a at x= -b/2a.
8. If α, β, γ are roots of cubic equation ax3 + bx2 + cx + d = 0, then, α + β + γ = -b/a, αβ + βγ + λα = c/a, and αβγ = -d/a