<em>Let the common root is ‘x’</em>
<em>Let the common root is ‘x’x2 + ax + b = 0 ……(1)</em>
<em>Let the common root is ‘x’x2 + ax + b = 0 ……(1)x2 + bx + a = 0 ……(2)</em>
<em>Let the common root is ‘x’x2 + ax + b = 0 ……(1)x2 + bx + a = 0 ……(2)Subtract equation (2) from (1), we get</em>
<em>Let the common root is ‘x’x2 + ax + b = 0 ……(1)x2 + bx + a = 0 ……(2)Subtract equation (2) from (1), we get(a – b)x + (b –a) = 0</em>
<em>Let the common root is ‘x’x2 + ax + b = 0 ……(1)x2 + bx + a = 0 ……(2)Subtract equation (2) from (1), we get(a – b)x + (b –a) = 0⇒ x = (a-b)/(a-b) = 1</em>
<em>Let the common root is ‘x’x2 + ax + b = 0 ……(1)x2 + bx + a = 0 ……(2)Subtract equation (2) from (1), we get(a – b)x + (b –a) = 0⇒ x = (a-b)/(a-b) = 1⇒ x = 1</em>
<em>Let the common root is ‘x’x2 + ax + b = 0 ……(1)x2 + bx + a = 0 ……(2)Subtract equation (2) from (1), we get(a – b)x + (b –a) = 0⇒ x = (a-b)/(a-b) = 1⇒ x = 1⇒ {1 + a + b = 0} (From equation (1))</em>
<em>Let the common root is ‘x’x2 + ax + b = 0 ……(1)x2 + bx + a = 0 ……(2)Subtract equation (2) from (1), we get(a – b)x + (b –a) = 0⇒ x = (a-b)/(a-b) = 1⇒ x = 1⇒ {1 + a + b = 0} (From equation (1))⇒ a + b = –1</em>
17-5=12
So Eric has 12 candy bars
The rate of change is 3 because each game costs $3 and the y-value will go up by 3 when the x-value goes up by 1. Hope this helps!
It would take him 7 (2/3) (seven and two-thirds) hours to read 230 pages.