Equations #17 - Questions and Answers

Question 1 Write the point-slope form of the line satisfying the given conditions. Then use the point-slope form of the equation to write the slope-intercept form of the equation. Slope = 6, passing through (-5,4). Answer Step 1 Given data is: Slope  𝑚 = 6 Slope m=6 Point  ( 𝑥 1 , 𝑦 1 ) = ( − 5 , 4 ) Point (x 1 ​ ,y 1 ​ )=(−5,4) The equation of line in point slope form is: 𝑦 − 𝑦 1 = 𝑚 ( 𝑥 − 𝑥 1 ) (1) y−y 1 ​ =m(x−x 1 ​ )(1) Substitute the value of 𝑚 m and ( 𝑥 1 , 𝑦 1 ) (x 1 ​ ,y 1 ​ ) in equation (1): 𝑦 − 4 = 6 ( 𝑥 − ( − 5 ) ) y−4=6(x−(−5)) 𝑦 − 4 = 6 ( 𝑥 + 5 ) y−4=6(x+5) Thus, the equation of the line in point-slope form is: 𝑦 − 4 = 6 ( 𝑥 + 5 ) y−4=6(x+5) Step 2 Equation of line in slope-intercept form from point-slope form: 𝑦 − 4 = 6 ( 𝑥 + 5 ) y−4=6(x+5) 𝑦 − 4 = 6 𝑥 + 30 y−4=6x+30 𝑦 = 6 𝑥 + 34 (2) y=6x+34(2) Standard form of slope-intercept equation is 𝑦 = 𝑚 𝑥 + 𝑐 y=mx+c. From standard equation of slope-intercept form, it's clear that equation (2) represents slope-intercept form. Thus, the equation of line in slope-intercept form is: 𝑦 = 6 𝑥 + 34 y=6x+34 Question 2 Solve each radical equation. Check potential solutions. 5 𝑥 = 𝑥 − 10 5x ​ =x−10 Answer Step 1 To solve the radical equation, isolate the radical expression and take square on both sides to solve the equation. Now the equation becomes quadratic after squaring each side, then solve the quadratic equation by factoring the terms. Check the potential solution; substitute back the two values of 𝑥 x in the radical equation to check which value of 𝑥 x satisfies the equation. Step 2 Isolate the radical expression from the equation: 5 𝑥 = 𝑥 − 10 5x ​ =x−10 Squaring each side to solve the equation: 5 𝑥 = ( 𝑥 − 10 ) 2 5x=(x−10) 2 5 𝑥 = 𝑥 2 − 20 𝑥 + 100 5x=x 2 −20x+100 𝑥 2 − 25 𝑥 + 100 = 0 x 2 −25x+100=0 Factorize: ( 𝑥 − 20 ) ( 𝑥 − 5 ) = 0 (x−20)(x−5)=0 𝑥 = 5 ,   𝑥 = 20 x=5,x=20 Now, check these two values of 𝑥 x by substituting back in the radical equation. Check 𝑥 = 5 x=5: 5 ( 5 ) = 5 − 10 5(5) ​ =5−10 25 = − 5 25 ​ =−5 5 ≠ − 5 5  =−5 Since 𝑥 = 5 x=5 does not satisfy the equation, it is not a solution. Check 𝑥 = 20 x=20: 5 ( 20 ) = 20 − 10 5(20) ​ =20−10 100 = 10 100 ​ =10 10 = 10 10=10 Since 𝑥 = 20 x=20 satisfies the equation, the solution is: 𝑥 = 20 x=20 Question 3 Find the equation for the line with the given properties. Express your answer using the general form or slope-intercept form of the equation. Parallel to the line 𝑥 = − 12 x=−12; containing the point (-5,3). Answer Step 1 Any line whose equation is of the form 𝑥 = 𝑎 x=a is a vertical line. Thus, the given line 𝑥 = − 12 x=−12 is a vertical line. Step 2 Any line parallel to the given line is also vertical. So the equation of the line that is parallel to the given line is of the form: 𝑥 = 𝑎 x=a This passes through ( − 5 , 3 ) (−5,3). This means 𝑥 = − 5 x=−5. Final Answer: The required line equation is: 𝑥 = − 5 x=−5 Question 4 Use the discriminant to determine whether each quadratic equation has two unequal real solutions, a repeated real solution (a double root), or no real solution, without solving the equation. 3 𝑥 2 + 5 𝑥 − 8 = 0 3x 2 +5x−8=0 Answer 1 Step 1 We know that for a standard quadratic equation 𝑎 𝑥 2 + 𝑏 𝑥 + 𝑐 = 0 ax 2 +bx+c=0, the discriminant 𝐷 D of the given function is calculated as: 𝐷 = 𝑏 2 − 4 𝑎 𝑐 D=b 2 −4ac If 𝐷 > 0 D>0, roots will be real and distinct in nature. If 𝐷 = 0 D=0, roots will be real and equal in nature. If 𝐷 < 0 D<0, roots will be imaginary in nature. Step 2 We have the given quadratic equation as: 3 𝑥 2 + 5 𝑥 − 8 = 0 3x 2 +5x−8=0 On comparing with the standard equation 𝑎 𝑥 2 + 𝑏 𝑥 + 𝑐 = 0 ax 2 +bx+c=0, we get 𝑎 = 3 , 𝑏 = 5 , 𝑐 = − 8 a=3,b=5,c=−8. Step 3 Calculate the discriminant: 𝐷 = 𝑏 2 − 4 𝑎 𝑐 D=b 2 −4ac 𝐷 = ( 5 ) 2 − 4 ( 3 ) ( − 8 ) D=(5) 2 −4(3)(−8) 𝐷 = 25 + 96 D=25+96 𝐷 = 121 D=121 Since 𝐷 > 0 D>0, the roots will be real and distinct. Final Answer: The quadratic equation 3 𝑥 2 + 5 𝑥 − 8 = 0 3x 2 +5x−8=0 has two unequal real solutions. Question 5 Solve the equation: 𝑒 2 𝑥 + 4 𝑒 𝑥 − 21 = 0 e 2x +4e x −21=0 Answer Step 1: A quadratic equation of the form 𝑧 2 + 𝑎 𝑧 + 𝑏 = 0 z 2 +az+b=0 can be factored as ( 𝑧 + 𝑎 ) ( 𝑧 + 𝑏 ) (z+a)(z+b). For a factored quadratic equation ( 𝑧 + 𝑎 ) ( 𝑧 + 𝑏 ) = 0 (z+a)(z+b)=0, the solution is 𝑧 = − 𝑎 , − 𝑏 z=−a,−b. Try to write the given equation as a quadratic equation. Step 2: Given equation is: 𝑒 2 𝑥 + 4 𝑒 𝑥 − 21 = 0 e 2x +4e x −21=0 Take 𝑧 = 𝑒 𝑥 z=e x . So the given equation becomes: 𝑧 2 + 4 𝑧 − 21 = 0 z 2 +4z−21=0 Factorize: 𝑧 2 + 7 𝑧 − 3 𝑧 − 21 = 0 z 2 +7z−3z−21=0 𝑧 ( 𝑧 + 7 ) − 3 ( 𝑧 + 7 ) = 0 z(z+7)−3(z+7)=0 ( 𝑧 − 3 ) ( 𝑧 + 7 ) = 0 (z−3)(z+7)=0 Step 3: Solve for 𝑧 z: 𝑧 = 3   or   𝑧 = − 7 z=3orz=−7 Since 𝑧 = 𝑒 𝑥 z=e x and 𝑒 𝑥 > 0 e x >0, discard 𝑧 = − 7 z=−7. 𝑒 𝑥 = 3 e x =3 ln ⁡ ( 𝑒 𝑥 ) = ln ⁡ ( 3 ) ln(e x )=ln(3) 𝑥 = ln ⁡ ( 3 ) 2 x= 2 ln(3) ​ Final Answer: 𝑥 = ln ⁡ ( 3 ) 2 x= 2 ln(3)