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Advanced Math / Nonlinear equations in one variable and systems of equations in two variables Difficulty: Hard

y+k=x+26

y-k=x2-5x

In the given system of equations, k is a constant. The system has exactly one distinct real solution. What is the value of k ?

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Explanation

The correct answer is 352. Subtracting the second equation from the first equation yields y+k-y-k=x+26-x2-5x, or 2k=-x2+6x+26. This is equivalent to x2-6x+2k-26=0. It's given that the system has exactly one distinct real solution; therefore, this equation has exactly one distinct real solution. An equation of the form ax2+bx+c=0, where a , b , and c are constants, has exactly one distinct real solution when the discriminant, b2-4ac, is equal to 0 . The equation x2-6x+(2k-26)=0 is of this form, where a=1, b=-6, and c=2k-26. Substituting these values into the discriminant, b2-4ac, yields -62-412k-26. Setting the discriminant equal to 0 yields -62-412k-26=0, or -8k+140=0. Subtracting 140 from both sides of this equation yields -8k=-140. Dividing both sides of this equation by - 8 yields k=352. Note that 35/2 and 17.5 are examples of ways to enter a correct answer.