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If 12p + 4q = 55, then p = (55 - 4q)/12

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7y ago
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7y ago

It is (55 - 4q)/12.

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Q: What is p when 12p plus 4q 55?
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What is 12p plus 19q plus p - 5q - 3p?

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If the ratio of the roots of the equation x2 plus px plus q0 is equal to the ratio of the roots of x62 plus lx plus m0 prove that mp2ql2.?

Thanks, Ankur. Let x^2+px+q=0 be Eqn 1 (I have shortened q0 to q) Let x^2+lx+m=0 be Eqn 2 I use the ^ to mean raised to the power of i.e. p^2 means p squared I use * to mean multiply The two roots of Eqn 1 (using the quadratic formula) are x =[-p±sqrt(p^2-4q)]/2 or separately [-p+sqrt(p^2-4q)]/2 and [-p-sqrt(p^2-4q)]/2 The two roots of Eqn 2 are x =[-l±sqrt(l^2-4m)]/2 or separately [-l+sqrt(l^2-4m)]/2 and [-l-sqrt(l^2-4m)]/2 Now we are told that the ratio of the roots of the two equations are equal. So {[-p+sqrt(p^2-4q)]/2} / {[-p-sqrt(p^2-4q)]/2} = {[-l+sqrt(l^2-4m)]/2} / {[-l-sqrt(l^2-4m)]/2} Cancel out the denominator 2 and cross multiply and ==>[-p+sqrt(p^2-4q)] * [ -l-sqrt(l^2-4m)] = [-p-sqrt(p^2-4q)] * [-l+sqrt(l^2-4m)] Simplify by multiplying each side ==>lp + p*sqrt(l^2-4m) - l*sqrt(p^2-4q) + [sqrt(p^2-4q)]*(l^2-4m)] = lp - p*sqrt(l^2-4m) + l*sqrt(p^2-4q) + [sqrt(p^2-4q)]*(l^2-4m)] Now cancel out lp as it's on both side and cancel out [sqrt(p^2-4q)]*(l^2-4m) because that's on both sides. Simplify by bring like terms together and ==>2p*sqrt(l^2-4m) = 2l*sqrt(p^2-4q) Cancel out the 2 on each side and square each side. ==>p^2(l^2-4m) = l^2(p^2-4q) Simply by multiplying out ==>p^2*l^2 - 4m*p^2 = p^2l^2 - l^2*4q Cancel out p^2*l^2 ==>-4m*p^2 = -4q l^2 Cancel out the -4 and voila ++> mp^2 = ql^2 If there is a shorter solution, I'd love to see it.


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9 Solve the inequality 12p plus 7 139?

The answer to 12 P plus 7 times 139 equals 39. This is taught in high school math.


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