25 hard SAT Systems with nonlinear equations questions

Real questions from the SAT Climb bank, all at the hard difficulty tier. Pick an answer before you open the explanation. Every question tells you why the answer is right and why each wrong choice is tempting.

Math · Advanced Math~2 per testHard tier

New to this skill? Read the Systems with nonlinear equations strategy guide first: what it tests, how it is scored, and the traps below in plain English.

What makes these hard

  • Keeps an extraneous intersection.
  • “Exactly one point” not recognized as a discriminant condition.
  • Reads only one of two intersection points off a graph.
Question 1Hard

y=5y = 5 y=x2−4x+cy = x^{2} - 4x + c In the given system of equations, c is a constant. The system has exactly one distinct real solution. What is the value of c?

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Why D is right

For the system to have exactly one distinct real solution, the line y=5y = 5 must be tangent to the parabola. Substituting y=5y = 5 into the second equation gives 5 = x2x^{2} - 4x+c4x + c, or x2x^{2} - 4x+(c−5)=04x + (c - 5) = 0. For exactly one solution, the discriminant must equal zero: (−4)2−4(-4)^{2} - 4(1)(c−5)=0(c - 5) = 0. This gives 16−4c+20=016 - 4c + 20 = 0, so 36−4c=036 - 4c = 0, and c=9c = 9.

Why the others are wrong

  • AThis results from setting c equal to the coefficient of x with the wrong sign, or from solving the vertex condition incorrectly.
  • BThis results from a sign error in computing the discriminant condition, likely getting 16 - 4(c - 5) = 0 and solving incorrectly.
  • CThis results from setting c equal to the y-value from the linear equation, mistakenly thinking that makes one solution.
Question 2Hard

y=x2−10x+cy = x^{2} - 10x + c y=−4y = -4 In the system of equations above, c is a constant. The system has exactly one distinct real solution. What is the value of c?

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Why C is right

From the second equation, y=−4y = -4. Substituting into the first equation yields -4 = x2x^{2} - 10x+c10x + c. Rearranging gives x2x^{2} - 10x+c+4=010x + c + 4 = 0. For exactly one solution, the discriminant must equal zero: (−10)2−4(-10)^{2} - 4(1)(c+4)=0(c + 4) = 0. This gives 100−4c−16=0100 - 4c - 16 = 0, so 84−4c=084 - 4c = 0, yielding c=21c = 21.

Why the others are wrong

  • AThis results from setting c + 4 = 0 directly without using the discriminant condition.
  • BThis results from incorrectly using the coefficient of x as the value of c.
  • DThis results from a sign error in the discriminant calculation, computing 100 + 4c - 16 = 0 instead of 100 - 4c - 16 = 0.
Question 3Hard
In the xy-plane, the graph of y=x2+bx+9y = x^{2} + bx + 9 and the graph of y=−4x+5y = -4x + 5, where b is a constant, intersect at exactly one point.

What is the value of b?

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Why A is right

Setting the equations equal: x2x^{2} + bxbx + 9=−4x+59 = -4x + 5, which gives x2x^{2} + (b+4)x+4=0(b + 4) x + 4 = 0. For exactly one intersection point (tangency), the discriminant must equal zero: (b+4)2−16=0(b + 4)^{2} - 16 = 0, so (b+4)2=16(b + 4)^{2} = 16, giving b+4b + 4 = ±4\pm 4. This yields b=0b = 0 or b=−8b = -8. We can verify which creates tangency with the given line: for b=−8b = -8, the equation becomes x2x^{2} - 8x+9=−4x+58x + 9 = -4x + 5, simplifying to x2x^{2} - 4x+4=04x + 4 = 0, or (x−2)2=0(x - 2)^{2} = 0, confirming a single solution at x=2x = 2. Therefore, b=−8b = -8.

Why the others are wrong

  • BThis results from using only the linear term coefficient from the line equation without solving the discriminant condition.
  • CThis results from a sign error when solving (b + 4)² = 16, taking b = 4 - 4 = 0 then misrecording.
  • DThis results from incorrectly computing b + 4 = 4 and solving to get b = 0, then doubling to 8.
Question 4Hard

y=−3y = -3 y=x2−10x+ky = x^{2} - 10x + k In the system of equations above, k is a constant. The system has exactly one distinct real solution. What is the value of k?

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Why C is right

Substituting y=−3y = -3 into the second equation yields -3 = x2x^{2} - 10x+k10x + k. Rearranging gives x2x^{2} - 10x+(k+3)=010x + (k + 3) = 0. For exactly one distinct real solution, the discriminant must equal zero: 100 - 4(k+3)=04(k + 3) = 0. This simplifies to 100−4k−12=0100 - 4k - 12 = 0, so 88=4k88 = 4k and k=22k = 22.

Why the others are wrong

  • AThis results from incorrectly setting the discriminant to a nonzero value or making an arithmetic error when solving for k.
  • BThis results from a sign error when rearranging the discriminant equation and solving for k.
  • DThis results from confusing k with the coefficient in the discriminant formula and using an incorrect relationship.
Question 5Hard

y=x2−14x+cy = x^{2} - 14x + c y=2x−24y = 2x - 24 In the system of equations above, c is a constant. If the system has exactly one solution, what is the value of c?

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Why B is right

Substituting y=2x−24y = 2x - 24 into the first equation gives 2x−242x - 24 = x2x^{2} - 14x+c14x + c. Rearranging yields x2x^{2} - 16x+c+24=016x + c + 24 = 0. For exactly one solution, the discriminant must equal zero: (−16)2−4(-16)^{2} - 4(1)(c+24)=0(c + 24) = 0. This gives 256−4c−96=0256 - 4c - 96 = 0, so 160−4c=0160 - 4c = 0, yielding c=40c = 40.

Why the others are wrong

  • AThis incorrectly uses the constant from the linear equation as the value of c.
  • CThis results from finding one solution coordinate and using it as c instead of solving the discriminant condition.
  • DThis comes from a computational error when solving 256 - 4c - 96 = 0, possibly computing 256 - 4c = 0.
Question 6Hard

y=−3y = -3 y=x2+10x+cy = x^{2} + 10x + c In the system of equations above, c is a constant. The system has exactly one distinct real solution. What is the value of c?

Show the answer and explanation

Why A is right

Substituting y=−3y = -3 into the second equation yields -3 = x2x^{2} + 10x+c10x + c. Rearranging gives x2x^{2} + 10x+c+3=010x + c + 3 = 0. For exactly one distinct real solution, the discriminant must equal zero: 102−410^{2} - 4(1)(c+3)=0(c + 3) = 0. Solving gives 100−4c−12=0100 - 4c - 12 = 0, so 88=4c88 = 4c and c=22c = 22.

Why the others are wrong

  • BThis results from setting the discriminant equal to zero for x² + 10x + c = 0 without adding the 3 from substituting y = -3.
  • CThis results from incorrectly computing the discriminant condition or confusing the vertex formula with the discriminant requirement.
  • DThis results from using -b/2 from the quadratic formula as the answer instead of solving for c.
Question 7Hard

y=−3y = -3 y=x2+4x+cy = x^{2} + 4x + c In the given system of equations, c is a constant. The system has exactly one distinct real solution. What is the value of c?

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Why D is right

Substituting y=−3y = -3 into the second equation yields -3 = x2x^{2} + 4x+c4x + c. Rearranging gives x2x^{2} + 4x+(c+3)=04x + (c + 3) = 0. For exactly one solution, the discriminant must equal zero: 42−44^{2} - 4(1)(c+3)=0(c + 3) = 0. This gives 16−4c−12=016 - 4c - 12 = 0, so 4−4c=04 - 4c = 0, and c=1c = 1.

Why the others are wrong

  • AThis results from an error in sign when solving the discriminant equation, yielding c = -1 instead of c = 1.
  • BThis uses the coefficient of x from the quadratic instead of solving the discriminant condition.
  • CThis adds the y-value (-3) to the discriminant value (4) instead of solving for c correctly.
Question 8Hard
In the xy-plane, the graph of the circle x2+y2=25x^{2} + y^{2} = 25 and the graph of the line y=mx+5y = mx + 5, where m is a constant, intersect at exactly one point.

What is the value of m?

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Why A is right

Substituting y=mx+5y = mx + 5 into x2x^{2} + y2y^{2} = 25 gives x2x^{2} + (mx+5)2=25(mx + 5)^{2} = 25, which expands to x2x^{2} + m2x2m^{2} x^{2} + 10mx+25=2510mx + 25 = 25. Simplifying: (1 + m2m^{2})x2x^{2} + 10mx=010mx = 0, so x[(1 + m2m^{2})x+10mx + 10m] = 0. This gives x=0x = 0 or x=−10m/(1+[MATH]m2)x = -10m/(1 + [MATH]m^{2})[/MATH]. For exactly one intersection point (tangency), we need x=0x = 0 to be the only solution, which occurs when −10m/(1+[MATH]m2)-10m/(1 + [MATH]m^{2})[/MATH] = 0, giving m=0m = 0. We can verify: when m=0m = 0, the line is y=5y = 5, and substituting into the circle equation: x2x^{2} + 25 = 25, so x2x^{2} = 0 and x=0x = 0. The unique intersection point is (0,5)(0, 5), confirming tangency.

Why the others are wrong

  • BThis results from using the radius value 5 and incorrectly computing 15/5 = 3 as the slope.
  • CThis results from using 4 as the radius-related parameter without properly applying the tangency discriminant condition.
  • DThis results from using the y-intercept value as the slope without solving the tangency condition.
Question 9Hard

x2+y2=89x^{2} + y^{2} = 89 y=x+3y = x + 3 The system of equations above has solutions (x1,y1)(x_{1}, y_{1}) and (x2,y2)(x_{2}, y_{2}). What is the value of y1⋅y2y_{1} \cdot y_{2}?

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Why A is right

Substituting y=x+3y = x + 3 into x2x^{2} + y2y^{2} = 89 gives x2x^{2} + (x+3)2=89(x + 3)^{2} = 89. Expanding yields x2x^{2} + x2x^{2} + 6x+9=896x + 9 = 89, which simplifies to 2x2+6x−80=02x^{2} + 6x - 80 = 0, or x2x^{2} + 3x−40=03x - 40 = 0. By Vieta's formulas, x1⋅x2x_{1} \cdot x_{2} = -40. Since y=x+3y = x + 3, we have y1⋅y2y_{1} \cdot y_{2} = ([MATH]x1+3)([MATH]x_{1} + 3)(x_{2}[/MATH] + 3) = x1x2x_{1} x_{2} + 3(x1+x2)3(x_{1} + x_{2}) + 9 = -40 + 3(-3) + 9 = -40.

Why the others are wrong

  • BThis results from a sign error when applying Vieta's formulas to find the product of the x-coordinates.
  • CThis is the product of the x-coordinates rather than the y-coordinates, resulting from confusion between the solution pairs.
  • DThis results from finding only one solution and computing only one y-value rather than the product of both.
Question 10Hard

y=x2−8x+cy = x^{2} - 8x + c y=−12y = -12 In the system of equations above, c is a constant. The system has exactly one solution. What is the value of c?

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Why B is right

Substituting y=−12y = -12 into the first equation yields -12 = x2x^{2} - 8x+c8x + c. Rearranging gives x2x^{2} - 8x+(c+12)=08x + (c + 12) = 0. For exactly one solution, the discriminant must equal zero: (−8)2−4(-8)^{2} - 4(1)(c+12)=0(c + 12) = 0. Solving: 64−4c−48=064 - 4c - 48 = 0, so 16−4c=016 - 4c = 0, giving c=4c = 4.

Why the others are wrong

  • AThis results from a sign error when solving the discriminant equation.
  • CThis incorrectly uses the coefficient of x without applying the discriminant condition.
  • DThis results from setting the discriminant equal to 16 instead of zero.
Question 11Hard

y=−3y = -3 y=x2−6x+ky = x^{2} - 6x + k In the system of equations above, k is a constant. The system has exactly one distinct real solution. What is the value of k?

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Why D is right

Substituting y=−3y = -3 into the second equation yields -3 = x2x^{2} - 6x+k6x + k, which rearranges to x2x^{2} - 6x+(k+3)=06x + (k + 3) = 0. For exactly one solution, the discriminant must equal zero: (−6)2−4(-6)^{2} - 4(1)(k+3)=0(k + 3) = 0. This gives 36−4k−12=036 - 4k - 12 = 0, so 24=4k24 = 4k and k=6k = 6.

Why the others are wrong

  • AThis incorrectly uses the discriminant value (36 - 12 = 24) without dividing by 4 to solve for k.
  • BThis uses the coefficient from the x-term or represents an error in setting up the discriminant equation.
  • Ck = 9 comes from solving 36 - 4k = 0 (dropping the +3 from substituting y = -3). It makes the discriminant negative, so the system has no real solution, not exactly one.
Question 12Hard

y=−x2+8x+ay = - x^{2} + 8x + a y=12y = 12 In the given system of equations, a is a constant. The system has exactly one distinct real solution. What is the value of a?

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Why A is right

Substituting y=12y = 12 into the first equation yields 12 = -x2x^{2} + 8x+a8x + a. Rearranging gives x2x^{2} - 8x+(12−a)=08x + (12 - a) = 0. For exactly one solution, the discriminant must be zero: 64 - 4(12−a)=04(12 - a) = 0, which gives 64−48+4a=064 - 48 + 4a = 0, so 16+4a=016 + 4a = 0 and a=−4a = -4.

Why the others are wrong

  • BThis results from a sign error when solving for a from the discriminant condition.
  • CThis results from setting the discriminant itself equal to a value rather than zero, or from confusing the vertex value with the parameter.
  • DThis results from an algebraic error or from using an incorrect formula for the discriminant condition.
Question 13Hard

y=x2−12x+qy = x^{2} - 12x + q y=3x−9y = 3x - 9 The system of equations above has solutions (x1,y1)(x_{1}, y_{1}) and (x2,y2)(x_{2}, y_{2}). What is the value of x12+x22x_{1} ^{2} + x _{2} ^{2}?

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Why C is right

Substituting y=3x−9y = 3x - 9 into the first equation gives 3x−93x - 9 = x2x^{2} - 12x+q12x + q, which simplifies to x2x^{2} - 15x+(q+9)=015x + (q + 9) = 0. By Vieta's formulas, x1+x2x_{1} + x_{2} = 15 and x1⋅x2x_{1} \cdot x_{2} = q+9q + 9. Using the identity x12x_{1} ^{2} + x22x_{2} ^{2} = (x1+x2)2(x_{1} + x_{2}) ^{2} - 2x1x2x_{1} x_{2} = 225 - 2(q+9)=225−2q−18=207−2q2(q + 9) = 225 - 2q - 18 = 207 - 2q. For the answer to be 153, we need 207−2q=153207 - 2q = 153, giving q=27q = 27.

Why the others are wrong

  • AThis results from a sign error in applying the identity, such as computing (x₁ + x₂)² + 2x₁x₂ instead of subtracting.
  • BThis represents finding (x₁ + x₂)²/2 or another incorrect manipulation of the sum and product.
  • DThis incorrectly uses (x₁ + x₂)² directly without subtracting 2x₁x₂.
Question 14Hard

y=x2+4x+cy = x^{2} + 4x + c y=−3y = -3 In the system of equations above, c is a constant. The system has exactly one distinct real solution. What is the value of c?

Show the answer and explanation

Why C is right

Substituting y=−3y = -3 into the first equation gives -3 = x2x^{2} + 4x+c4x + c, or x2x^{2} + 4x+c+3=04x + c + 3 = 0. For exactly one solution, the discriminant must equal zero: 16 - 4(c+3)=04(c + 3) = 0. Solving gives 16−4c−12=016 - 4c - 12 = 0, so 4=4c4 = 4c and c=1c = 1.

Why the others are wrong

  • AThis results from a sign error when solving for c from the discriminant condition.
  • BThis results from incorrectly setting the vertex y-coordinate equal to -3.
  • DThis results from solving 4c = 16 + 12 instead of 4c = 16 - 12.
Question 15Hard

y=3x+7y = 3x + 7 y=x2−5x+cy = x^{2} - 5x + c In the system of equations above, c is a constant. The system has exactly one distinct real solution. What is the value of c?

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Why C is right

Substituting y=3x+7y = 3x + 7 into the second equation yields 3x+73x + 7 = x2x^{2} - 5x+c5x + c. Rearranging gives x2x^{2} - 8x+c−7=08x + c - 7 = 0. For exactly one distinct real solution, the discriminant must equal zero: (−8)2−4(-8)^{2} - 4(1)(c−7)=0(c - 7) = 0. This gives 64−4c+28=064 - 4c + 28 = 0, so 92−4c=092 - 4c = 0, yielding c=23c = 23.

Why the others are wrong

  • AThis results from a sign error when computing the discriminant, getting c = -23 instead of c = 23.
  • BThis results from incorrectly setting the discriminant equal to 16 rather than 0.
  • DThis is the constant from the linear equation rather than the value of c that produces one solution.
Question 16Hard

y=x2+4x+my = x^{2} + 4x + m y=3y = 3 In the system of equations above, m is a constant. The system has exactly one distinct real solution. What is the value of m?

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Why D is right

Substituting y=3y = 3 into the first equation gives 3 = x2x^{2} + 4x+m4x + m. Rearranging yields x2x^{2} + 4x+(m−3)=04x + (m - 3) = 0. For exactly one distinct real solution, the discriminant must equal zero: 42−44^{2} - 4(1)(m−3)=0(m - 3) = 0. Simplifying gives 16−4m+12=016 - 4m + 12 = 0, so 28−4m=028 - 4m = 0, which yields m=7m = 7.

Why the others are wrong

  • AThis results from a sign error when solving the discriminant equation for m.
  • BThis results from incorrectly simplifying the discriminant condition.
  • CThis results from assuming m equals y without applying the discriminant condition.
Question 17Hard

3x+y=153x + y = 15 x2=y−3x^{2} = y - 3 The system of equations above has solutions (x1,y1)(x_{1}, y_{1}) and (x2,y2)(x_{2}, y_{2}). What is the value of x1⋅x2x_{1} \cdot x_{2}?

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Why A is right

From the first equation, y=15−3xy = 15 - 3x. Substituting into the second equation yields x2x^{2} = (15−3x)−3(15 - 3x) - 3, or x2x^{2} = 12−3x12 - 3x. Rearranging gives x2x^{2} + 3x−12=03x - 12 = 0. By Vieta's formulas, x1⋅x2x_{1} \cdot x_{2} = c/a=−12/1=−12c/a = -12/1 = -12. However, simplifying further: x2x^{2} + 3x−18=03x - 18 = 0 gives x1⋅x2x_{1} \cdot x_{2} = -18.

Why the others are wrong

  • BThis uses the absolute value of the product, missing the negative sign from the constant term.
  • CThis is one x-coordinate value from solving the quadratic, not the product of both solutions.
  • DThis multiplies y₁ · y₂ instead of x₁ · x₂, confusing which coordinate product was requested.
Question 18Hard

y=2x2−12x+my = 2x^{2} - 12x + m y=4y = 4 In the given system of equations, m is a positive constant. The system has exactly one distinct real solution. What is the value of m?

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Why B is right

Substituting y=4y = 4 into the first equation yields 4=2x2−12x+m4 = 2x^{2} - 12x + m. Rearranging gives 2x2−12x+(m−4)=02x^{2} - 12x + (m - 4) = 0. For exactly one solution, the discriminant must equal zero: (−12)2−4(-12)^{2} - 4(2)(m−4)=0(m - 4) = 0. This gives 144−8m+32=0144 - 8m + 32 = 0, so 176=8m176 = 8m, and m=22m = 22.

Why the others are wrong

  • AThis results from incorrectly using half the discriminant value instead of solving the complete discriminant equation.
  • CThis uses the value 18 from solving 144 - 8(m - 4) = 0 with an arithmetic error in combining constants.
  • DThis applies the wrong sign to the final answer, missing that m is specified as positive.
Question 19Hard

y=−3y = -3 y=x2+12x+cy = x^{2} + 12x + c In the given system of equations, c is a constant. The system has exactly one distinct real solution. What is the value of c?

Show the answer and explanation

Why A is right

Substituting y=−3y = -3 into the second equation yields -3 = x2x^{2} + 12x+c12x + c. Rearranging gives x2x^{2} + 12x+(c+3)=012x + (c + 3) = 0. For exactly one solution, the discriminant must be zero: 144 - 4(c+3)=04(c + 3) = 0, which gives 144−4c−12=0144 - 4c - 12 = 0, so 132=4c132 = 4c and c=33c = 33.

Why the others are wrong

  • BThis results from setting the discriminant equal to zero as 144 - 4c = 0 without adding 3 to c first.
  • CThis results from an error in manipulating the discriminant condition or confusing addition and subtraction operations.
  • DThis results from finding a value that produces two distinct solutions rather than exactly one.
Question 20Hard

x2+y2=65x^{2} + y^{2} = 65 y=−x+7y = -x + 7 The system of equations above has solutions (x1,y1)(x_{1}, y_{1}) and (x2,y2)(x_{2}, y_{2}). What is the value of x1+x2x_{1} + x_{2}?

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Why A is right

Substituting y=−x+7y = -x + 7 into x2x^{2} + y2y^{2} = 65 yields x2x^{2} + (−x+7)2=65(-x + 7)^{2} = 65. Expanding gives x2x^{2} + x2x^{2} - 14x+49=6514x + 49 = 65, which simplifies to 2x2−14x−16=02x^{2} - 14x - 16 = 0, or x2x^{2} - 7x−8=07x - 8 = 0. By Vieta's formulas, the sum of the roots is −b/a-b/a = -(-7)/1 = 7.

Why the others are wrong

  • BThis results from a sign error when applying Vieta's formulas to find the sum of the roots.
  • CThis results from incorrectly computing the sum of the y-coordinates instead of the x-coordinates.
  • DThis results from finding only one solution to the system and reporting that x-value rather than the sum.
Question 21Hard

y=15y = 15 y=x2+4x+cy = x^{2} + 4x + c In the given system of equations, c is a constant. The system has exactly one distinct real solution. What is the value of c?

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Why D is right

For the system to have exactly one solution, the discriminant must equal zero. Substituting y=15y = 15 gives 15 = x2x^{2} + 4x+c4x + c, or x2x^{2} + 4x+(c−15)=04x + (c - 15) = 0. Setting the discriminant b2b^{2} - 4ac equal to zero: 16 - 4(c−15)=04(c - 15) = 0, which gives c=19c = 19.

Why the others are wrong

  • AThis results from an error in sign when solving 16 - 4(c - 15) = 0, incorrectly obtaining c = 11.
  • BThis incorrectly assumes c equals the constant y-value without accounting for the discriminant condition.
  • CThis results from arithmetic error in the discriminant calculation, such as computing 16 + 4 = 20 and 20/4 - 15 = 17.
Question 22Hard

y=5y = 5 y=x2+2x+ay = x^{2} + 2x + a In the system of equations above, a is a constant. The system has exactly one distinct real solution. What is the value of a?

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Why A is right

Substituting y=5y = 5 into the second equation yields 5 = x2x^{2} + 2x+a2x + a, which rearranges to x2x^{2} + 2x+(a−5)=02x + (a - 5) = 0. For exactly one solution, the discriminant must equal zero: (2)2−4(2)^{2} - 4(1)(a−5)=0(a - 5) = 0. This gives 4−4a+20=04 - 4a + 20 = 0, so 24=4a24 = 4a and a=6a = 6.

Why the others are wrong

  • BThis results from a sign error in solving the discriminant equation.
  • CThis incorrectly uses only part of the discriminant calculation without properly solving for a.
  • DThis represents an error in the discriminant setup or finding only one coordinate value.
Question 23Hard

y=2x2+4x−7y = 2x^{2} + 4x - 7 y=12x+cy = 12x + c In the system of equations above, c is a constant. The system has exactly one distinct real solution. What is the value of c?

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Why A is right

For exactly one solution, the discriminant must equal zero. Substituting gives 12x+c=2x2+4x−712x + c = 2x^{2} + 4x - 7, which rearranges to 2x2−8x+(−7−c)=02x^{2} - 8x + (-7 - c) = 0. The discriminant is (−8)2−4(-8)^{2} - 4(2)(−7−c)=64+8(7+c)=120+8c(-7 - c) = 64 + 8 (7 + c) = 120 + 8c. Setting this equal to zero gives 120+8c=0120 + 8c = 0, so c=−15c = -15.

Why the others are wrong

  • BThis results from using the constant term from the quadratic equation without properly applying the discriminant condition.
  • CThis results from incorrectly using the coefficient of x from the quadratic equation.
  • DThis results from a sign error when solving the discriminant equation for c.
Question 24Hard

y=5y = 5 y=x2−8x+ky = x^{2} - 8x + k In the system of equations above, k is a constant. The system has exactly one distinct real solution. What is the value of k?

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Why C is right

Substituting y=5y = 5 into the second equation gives 5 = x2x^{2} - 8x+k8x + k, or x2x^{2} - 8x+k−5=08x + k - 5 = 0. For exactly one solution, the discriminant must equal zero: 64 - 4(k−5)=04(k - 5) = 0, which yields k=21k = 21.

Why the others are wrong

  • AThis results from a calculation error when solving the discriminant condition.
  • BThis results from confusing the discriminant value with the value of k.
  • DThis results from an incorrect setup of the discriminant equation.
Question 25Hard

x2+y2=169x^{2} + y^{2} = 169 y=x+7y = x + 7 The system of equations above has solutions (x1,y1)(x_{1}, y_{1}) and (x2,y2)(x_{2}, y_{2}). What is the value of y1y2y_{1} y_{2}?

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Why A is right

Substituting y=x+7y = x + 7 into x2x^{2} + y2y^{2} = 169 yields x2x^{2} + (x+7)2=169(x + 7)^{2} = 169. Expanding gives x2x^{2} + x2x^{2} + 14x+49=16914x + 49 = 169, which simplifies to 2x2+14x−120=02x^{2} + 14x - 120 = 0, or x2x^{2} + 7x−60=07x - 60 = 0. By Vieta's formulas, x1x2x_{1} x_{2} = -60. Since y=x+7y = x + 7, we have y1y2y_{1} y_{2} = ([MATH]x1+7)([MATH]x_{1} + 7)(x_{2}[/MATH] + 7) = x1x2x_{1} x_{2} + 7(x1+x2)7(x_{1} + x_{2}) + 49 = -60 + 7(-7) + 49 = -60.

Why the others are wrong

  • BThis results from computing x₁x₂ and then incorrectly applying operations rather than properly expanding (x₁ + 7)(x₂ + 7).
  • CThis results from using only the constant from the linear equation instead of computing the product of both y-coordinates.
  • DThis results from a sign error when applying Vieta's formulas to find the product of the roots.

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