a) The value of a, b and c for the quadratic expressions are as follows;
1) a =1, b = -4, c = 5
2) a =2, b = 1, c = -6
3) a =1, b = -3, c = 1
4) a =1, b = -4, c = -7
5) a = -3, b = 2, c = 8
The standard formula of a quadratic equation is expressed as:
[tex]ax^2+bx+c=0[/tex]
where a, b and c are constants
Given the following quadratic expressions, we are to find the constants a, b and c.
1) For the expression x² − 4x− 5 = 0
Comparing with the general formula
ax² = x²
a = 1
Find b
-4x = bx
b = -4
Find c
c = -5
2) For the expression 2x² + x− 6 = 0
Comparing with the general formula
ax² = 2x²
a = 2
Find b
x = bx
b = 1
Find c
c = -6
3) For the expression x² − 3x+ 1 = 0
Comparing with the general formula
ax² = x²
a = 1
Find b
-3x = bx
b = -3
Find c
c = 1
4) For the expression x² − 6x + 2x - 7 = 0
x² − 4x - 7 = 0
Comparing with the general formula
ax² = x²
a = 1
Find b
-4x = bx
b = -4
Find c
c = -7
4) For the expression − 2x = -3x² + 8
-3x² + 2x + 8 = 0
Comparing with the general formula
ax² = -3x²
a = -3
Find b
2x = bx
b = 2
Find c
c = 8
b) The general quadratic formula is expressed as:
[tex]x=\frac{-b\pm\sqrt{b^2-4ac} }{2a}[/tex]
1) To solve the quadratic expressions in (a), we only need to substitute the constants into the formula
[tex]x=\frac{-(-4)\pm\sqrt{(-4)^2-4(1)(-5)} }{2(1)}\\x=\frac{4\pm\sqrt{(16+20} }{2(1)}\\x=\frac{4\pm\sqrt{36} }{2}\\x=\frac{4+6}{2} \ and \ \frac{4-6}{2} \\x=5 \ and \ -1[/tex]
2) 2x² + x− 6 = 0
[tex]x=\frac{-1\pm\sqrt{(1)^2-4(2)(-6)} }{2(2)}\\x=\frac{-1\pm\sqrt{(1+48} }{4}\\x=\frac{-1\pm 7 }{2}\\x=\frac{-1+7}{2} \ and \ \frac{-1-7}{2} \\x=3 \ and \ -4[/tex]
3) x² − 3x + 1 = 0
[tex]x=\frac{3\pm\sqrt{(-3)^2-4(1)(1)} }{2(1)}\\x=\frac{3\pm\sqrt{9-4} }{2}\\x=\frac{3\pm\sqrt{(9-4} }{2}\\x=\frac{3+ \sqrt{5} }{2} \ and \ x=\frac{3-\sqrt{5} }{2}[/tex]
4) For x² − 4x - 7 = 0
[tex]x=\frac{4\pm\sqrt{(4)^2-4(1)(-7)} }{2(1)}\\x=\frac{4\pm\sqrt{16+28} }{2}\\x=\frac{4\pm\sqrt{44} }{2}\\x=\frac{4+ \sqrt{44} }{2} \ and \ x=\frac{4-\sqrt{44} }{2}[/tex]
5) For the expression -3x² + 2x + 8 = 0
[tex]x=\frac{-2\pm\sqrt{(2)^2-4(-3)(8)} }{2(-3)}\\x=\frac{-2\pm\sqrt{4+96} }{-6}\\x=\frac{-2\pm\sqrt{100} }{-6}\\x=\frac{-2\pm 10} {-6}\\x=\frac{-2+10}{-6} \ and \ \frac{-2-10}{-6}\\x=\frac{-8}{6} \ and \ x = -2\\x = \frac{-4}{3} \ and \ x = -2[/tex]
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