To solve: 4t25=t5+310

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Chanell Sanborn · Accepted Answer

Step 1  Given equation: 4t25=t5+310  Multiply by 10 on both sides,  10×4t25=(t5+310)10  By distributive property,  (t5+310)10=10×t5+10×310  (t5+310)10=2t+3  10×4t25=8t2  8t2=2t+3  Subtract −2t−3 on both sides,  8t2−2t−3=2t+3−2t−3  8t2−2t−3=0  Split the middle term −2t as −6t and +4t  Replace −2t with −6t+4t in 8t2−2t−3=0  8t2−6t+4t−3=0  Step 2  The above equation can be written as 2t×4t−2t×3+1×4t−3×1  Apply distributive property for first two terms and last two terms,  2t×4t−2t×3+1×4t−3×1=2t(4t−3)+1(4t−3)  Again apply distributive property for 2t(4t−3)+1(4t−3)  2t(4t−3)+1(4t−3)=(2t+1)(4t−3)  So the factors are (2t+1) and (4t−3)  8t2−2t−3=(2t+1)(4t−3)  To find the value of t,  Take 4t−3=0  Add 3 on both sides,  4t−3+3=0+3  Add the like terms,  4t=3  Divide by 4 on both sides,  4t4=34

Durst37 · Accepted Answer

Step 1  Find Least Common Multiplier of 5, 10: 10  Multiply by LCM=10  4t25×10=t5×10+310×10  Simplify  8t2=2t+3  Subtract 3 from both sides  8t2−3=2t+3−3  Simplify  8t−2t−3=0  Solve with the quadratic formula  t1, 2=−(−2)±(−2)2−4×8(−3)2×8  (−2)2−4×8(−3)=10  t1, 2=−(−2)±102×8  Separate the solutions  t1=−(−2)+102×8  t2=−(−2)−102×8  t1= Apply rule −(a)=a  =2+102×8  Add the numbers: 2+10=12  =122×8  Multiply the numbers: 2×8=16  =1216  Cancel the common factor: 4  =34  t2=  =2−102×8  2−10=−8  =−82×8  2×8=16  =−816  Apply the fraction rule: −ab=−ab  =−816  Cancel the common factor: 8  =−12  The solutions to the quadratic equation are:  t=34, t=−12

karton · Accepted Answer

Given: 4t25=t5+310 Multiply both sides of the equation by 10, the least common multiple of 5,10 2×4t2=2t+3 Multiply 2 and 4 to get 8. 8t2=2t+3 Subtract 2t from both sides. 8t2−2t=3 Subtract 3 from both sides. 8t2−2t−3=0 To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 8t2+at+bt−3 To find a and b, set up a system to be solved. a+b=-2 ab=8(-3)=-24 Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -24. 1, -24 2, -12 3, -8 4, -6 Calculate the sum for each pair. 1-24=-23 2-12=-10 3-8=-5 4-6=-2 The solution is the pair that gives sum -2. a=-6 b=4 Rewrite 8t2−2t−3 as (8t2−6t)+(4t−3) (8t2−6t)+(4t−3) Factor out 2t in 8t2−6t 2t(4t-3)+4t-3 Factor out common term 4t-3 by using distributive property. (4t-3)(2t+1) To find equation solutions, solve 4t-3=0 and 2t+1=0. t=34 t=−12

To solve: \frac{4t^{2}}{5}=\frac{t}{5}+\frac{3}{10}

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