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Worked solution

Quadratics positive for every real input

(a) Show that if x2>k(x+1)x^2>k(x+1) for all real xx, then −4<k<0-4<k<0.

(b) Find the range of values of kk for which x2+2kx+15−2kx^2+2kx+15-2k is always positive for all real xx. Deduce the range of values of kk for which k(x2+2kx+15−2k)>0k(x^2+2kx+15-2k)>0 for all real xx.

CJC · Discussion · Question 6

Answer

(a) −4<k<0-4<k<0. (b) −5<k<3-5<k<3; product positive for all real inputs when 0<k<30<k<3.

(a)

x2−kx−k>0x^2-kx-k>0 for all real xx. The leading coefficient is positive, so the discriminant must be negative: k2+4k<0k^2+4k<0, or k(k+4)<0k(k+4)<0. Hence −4<k<0-4<k<0.

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(b)

The discriminant condition is 4k2−60+8k<04k^2-60+8k<0, hence k2+2k−15=(k+5)(k−3)<0k^2+2k-15=(k+5)(k-3)<0. The sign chart is positive, negative, positive across −5-5 and 33, giving −5<k<3-5<k<3.

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Deduction

For the product to be positive, either the quadratic and k are both negative or both positive. The quadratic cannot be negative for every real x because its leading coefficient is positive. Thus require k>0k>0 and −5<k<3-5<k<3. Their intersection is 0<k<30<k<3.

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The expression (k+5)(k−3) is negative between its excluded roots −5 and 3. Intersect this interval with k>0 to obtain 0<k<3.
The expression (k+5)(k−3) is negative between its excluded roots −5 and 3. Intersect this interval with k>0 to obtain 0<k<3.
The expression (k+5)(k−3) is negative between its excluded roots −5 and 3. Intersect this interval with k>0 to obtain 0<k<3.

Common mistakes

A zero discriminant allows a zero value, so it fails strict positivity.

Solution notes

The solution diagrams are independently drawn from the printed mathematical relationships. Geometry is illustrative; use the labelled critical values and endpoint symbols.

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