<span>Let p, np be the roots of the given QE.So p+np = -b/a, and np^2 = c/aOr (n+1)p = -b/a or p = -b/a(n+1)So n[-b/a(n+1)]2 = c/aor nb2/a(n+1)2 = cor nb2 = ac(n+1)2
Which will give can^2 + (2ac-b^2)n + ac = 0, which is the required condition.</span>
Answer:
103
Step-by-step explanation:
it has to add up to 180 so
180-77=103
Answer:
7x² + 4x + 32
Step-by-step explanation:
13x² + 7x +21
<u>- 6x² + 3x - 11 </u>
7x² +4x + 32
Answer:
Step-by-step explanation:
not sure sorry ask a parent
Answer:
(x - 10)(x + 10)
Step-by-step explanation:
x² - 100 ← is a difference of squares and factors in general as
a² - b² = (a - b)(a + b)
Hence
x² - 100
= x² - 10² = (x - 10)(x + 10)
