Further Pure Core 1 Watch

tpitt549
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The quadratic equation ax^2+ bx + c = 0 has two real roots in the ratio m : n.

(i) Show that ac(m + n)^2= mnb^2

Any ideas how to answer??
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ghostwalker
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(Original post by tpitt549)
The quadratic equation ax^2+ bx + c = 0 has two real roots in the ratio m : n.

(i) Show that ac(m + n)^2= mnb^2

Any ideas how to answer??
One way.

If we call one root alpha, then the other is...?

Sum of roots (-b/a) is.... Product of roots is .... Both in terms of alpha , m, n.

Then start with the LHS, using the derived information.

Bit messy, and I have a suspicion there's a more elegant method, but it illudes me at present.

Further hint on derivation in spoiler.

Spoiler:
Show

Note ac = a^2 (c/a)
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Prasiortle
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(Original post by tpitt549)
The quadratic equation ax^2+ bx + c = 0 has two real roots in the ratio m : n.

(i) Show that ac(m + n)^2= mnb^2

Any ideas how to answer??
Let the roots be mk and nk. By Vieta's Formulae, mk+nk = -b/a and mk*nk = c/a.
The first equation becomes (m+n)k = -b/a, so m+n = -b/(ak). Thus ac(m+n)^2 = ac*b^2/(a^2*k^2) = (cb^2)/(ak^2). From the second equation, mnk^2 = c/a, so mn = c/(ak^2). Thus from what we just had, ac(m+n)^2 = (cb^2)/(ak^2) = (c/ak^2) * b^2 = mn * b^2 = mnb^2, as required.
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ghostwalker
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(Original post by Prasiortle)
...
Please don't post full solutions straight off - see forum guidelines, sticky thread at top of forum.
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tpitt549
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(Original post by ghostwalker)
One way.

If we call one root alpha, then the other is...?

Sum of roots (-b/a) is.... Product of roots is .... Both in terms of alpha , m, n.

Then start with the LHS, using the derived information.

Bit messy, and I have a suspicion there's a more elegant method, but it illudes me at present.

Further hint on derivation in spoiler.

Spoiler:
Show


Note ac = a^2 (c/a)

Thank You!!! Took me a while to work out but useful knowing where to start! Now I can freak out slightly less about my Pure Mock tommorow!
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old_engineer
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(Original post by tpitt549)
The quadratic equation ax^2+ bx + c = 0 has two real roots in the ratio m : n.

(i) Show that ac(m + n)^2= mnb^2

Any ideas how to answer??
Just to offer one more variant, you can find the two roots in terms of a, b and c via the quadratic root formula, call them m and n, put one over the other and rearrange. It’s a bit messy. Isolating ac is the key.
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