Answer:
<h2>
∠PQT = 72°</h2>
Step-by-step explanation:
According to the diagram shown, ∠OPQ = ∠OQP = 18°. If PQT is a tangent to the circle, it can be inferred that line OQ is perpendicular to line QT. Ths shows that ∠OQT = 90°.
Also from the diagram, ∠OQP + ∠PQT = ∠OQT;
∠PQT = ∠OQT - ∠OQP
Given ∠OQP = 18° and ∠OQT = 90°
∠PQT = 90°-18°
∠PQT = 72°
<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>
Quadratic Equation in Standard Form: ax2 + bx + c = 0.
Quadratic Equations can be factored.
Quadratic Formula: x = −b ± √(b2 − 4ac) 2a.
When the Discriminant (b2−4ac) is: positive, there are 2 real solutions. zero, there is one real solution. negative, there are 2 complex solutions.
HOPE THIS HELPS!!!
